Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 1

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 2

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 3

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 4

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 5

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 6

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 7

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 8

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 9

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 10

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 11

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 12

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 13

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 14

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 15

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 16

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 17

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 18

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 19

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 20

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 21

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 22

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 23

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 24

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 25

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 26

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 27

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 28

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 29

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 30

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 31

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 32

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 33

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 34

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 35

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 36

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 37

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 38

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 39

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 40

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 41

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 42

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 43

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 44

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 45

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 46

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 47

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)  

eSolutions Manual - Powered by Cognero Page 48

10-2 Arithmetic Sequences and Series

Find the indicated term of each arithmetic sequence. 

1. a1 = 14, d = 9, n = 11

 

SOLUTION:  

 

2. a18 for 12, 25, 38, …

 

SOLUTION:  25 – 12 = 13 38 – 25 = 13   The common difference d is 13.  a1

= 12

 

 

Write an equation for the nth term of each arithmetic sequence.

  3. 13, 19, 25, …  

SOLUTION:  19 – 13 = 6 25 – 19 = 6   The common difference d is 6.  a1 = 13

 

 

4. a5 = –12, d = –4

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

Find the arithmetic means in each sequence. 

5. 

 

SOLUTION:  

Here a1 = 6 and a5 = 42.

 

  Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 

6. 

 

SOLUTION:  

Here a1 = –4 and a5 = 8.

 

  Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5. 

Find the sum of each arithmetic series. 

7. the first 50 natural numbers 

SOLUTION:  

 

8. 4 + 8 + 12 + … + 200  

SOLUTION:  8 – 4 = 4 12 – 8 = 4   The common difference is 4. 

 

  Find the sum of the series.  

 

9. a1 = 12, an = 188, d = 4

 

SOLUTION:  

  Find the sum of the series. 

 

10. an = 145, d = 5, n = 21

 

SOLUTION:  

  Find the sum of the series. 

 

Find the first three terms of each arithmetic series. 

11. a1 = 8, an = 100, Sn = 1296

 

SOLUTION:  

 

  Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 

12. n = 18, an = 112, Sn = 1098

 

SOLUTION:  

 

  Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22. 

13. MULTIPLE CHOICE Find

  A 45   B 78   C 342   D 410  

SOLUTION:  There are 12 – 1 + 1 or 12 terms, so n = 12. 

  Find the sum.  

  Option C is the correct answer.  

Find the indicated term of each arithmetic sequence. 

14. a1 = –18, d = 12, n = 16

 

SOLUTION:  

 

15. a1 = –12, n = 66, d = 4

 

SOLUTION:  

 

16. a1 = 9, n = 24, d = –6

 

SOLUTION:  

 

17. a15 for –5, –12, –19, …

 

SOLUTION:  –12 – (–5) = –7 –19 – (–12) = –7   The common difference d is –7.  a1

= –5

 

 

18. a10 for –1, 1, 3, …

 

SOLUTION:  1 – (–1) = 2 3 – 1 = 2   The common difference d is 2.  a1

= –1

 

 

19. a24 for 8.25, 8.5, 8.75, …

 

SOLUTION:  8.5 – 8.25 = 0.25 8.75 – 8.5 = 0.25   The common difference d is 0.25.  a1

= 8.25

 

 

Write an equation for the nth term of each arithmetic sequence.

  20. 24, 35, 46, …

 

SOLUTION:  35 – 24 = 11 46 – 35 = 11   The common difference d is 11.  a1 = 24

 

 

21. 31, 17, 3, …  

SOLUTION:  17 – 31 = –14 3 – 17 = –14   The common difference d is –14.  a1

= 31

 

 

22. a9 = 45, d = –3

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

23. a7 = 21, d = 5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

24. a4 = 12, d = 0.25

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

25. a5 = 1.5, d = 4.5

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

26. 9, 2, –5, …  

SOLUTION:  2 – 9 = –7 –5 – 2 = –7   The common difference d is –7.  a1 = 9

 

 

27. a6 = 22, d = 9

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

28. a8 = –8, d = –2

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

29. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

30. –12, –17, –22, …  

SOLUTION:  –17 – (–12) = –5 –22 – (–17) = –5   The common difference d is –5.  a1 = –12

 

 

31. 

 

SOLUTION:  

  Use the value of a1 to find the nth term.

 

 

32. CCSS STRUCTURE  José averaged 123 total pins per game in his bowing league this season. He is taking bowling lessons and hopes to bring his average up by 8 pins each new season.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during what season will José average 187 per game?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given d = 8 and a1 = 123.

Find the nth term.

  b. Substitute 187 for an and solve for n.

Therefore José’s average will be 187 pins per game in the 9th

season.

  c. Sample answer: No; there are a maximum of 300 points in a bowling game, so it would be impossible for the average to continue to climb indefinitely.  

Find the arithmetic means in each sequence. 

33. 

 

SOLUTION:  

Here a1 = 24 and a6 = –1.

 

  Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4. 

34.  ]

 

SOLUTION:  

Here a1 = –6 and a6 = 49.

 

  Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38. 

35. 

 

SOLUTION:  

Here a1 = –28 and a6 = 7.

 

  Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0. 

36. 

 

SOLUTION:  

Here a1 = 84 and a6 = 39.

 

  Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 

37. 

 

SOLUTION:  

Here a1 = –12 and a7 = –66.

 

  Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48 – 9) or –57.  

38. 

 

SOLUTION:  

Here a1 = 182 and a7 = 104.

 

  Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) or 117.  

Find the sum of each arithmetic series. 

39. the first 100 even natural numbers 

SOLUTION:  

Here a1 = 2 and a100 = 200.

  Find the sum.  

 

40. the first 200 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a200 = 399.

  Find the sum.  

 

41. the first 100 odd natural numbers 

SOLUTION:  

Here a1 = 1 and a100 = 199.

  n = 100   Find the sum.  

 

42. the first 300 even natural numbers 

SOLUTION:  

Here a1 = 2 and a300 = 300.

  n = 300   Find the sum.  

 

43. –18 + (–15) + (–12) + … + 66 

SOLUTION:  –15 – (–18) = 3 –12 – (–15) = 3   The common difference is 3. 

  Find the value of n.  

  Find the sum.  

 

44. –24 + (–18) + (–12) + … + 72 

SOLUTION:  –18 – (–24) = 6 –12 – (–18) = 6   The common difference is 6. 

  Find the value of n.  

  Find the sum.  

 

45. a1 = –16, d = 6, n = 24

 

SOLUTION:  

 

46. n = 19, an = 154, d = 8

 

SOLUTION:  

Find the value of a1.

  Find the sum.  

 

47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?  

SOLUTION:  

Given, a1 = 150, d = 50 and n = 11.

  Find the value of a11.

 

  Find the sum.  

  A cash prizes totaled $4400 for the eleven week contest. 

Find the first three terms of each arithmetic series. 

48. n = 32, an = –86, Sn = 224

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  So, the sequence is 100, 94, 88, … 

49. a1 = 48, an = 180, Sn = 1368

 

SOLUTION:  Find the value of n.

  Find the value of d.

  So, the sequence is 48, 60, 72, … 

50. a1 = 3, an = 66, Sn = 759

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are 3, 6 and 9. 

51. n = 28, an = 228, Sn = 2982

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –15, –6 and 3. 

52. a1 = –72, an = 453, Sn = 6858

 

SOLUTION:  Find the value of n.  

  Find the value of d.  

  Therefore, the first three terms are –72, –57 and –42. 

53. n = 30, an = 362, Sn = 4770

 

SOLUTION:  

Find the value of a1.

 

  Find the value of d.  

  Therefore, the first three terms are –44, –30 and –16. 

54. a1 = 19, n = 44, Sn = 9350

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are 19, 28 and 37. 

55. a1 = –33, n = 36, Sn = 6372

 

SOLUTION:  

Find the value of an.

 

  Find the value of d.  

  Therefore, the first three terms are –33, –21 and –9. 

56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.  

SOLUTION:  

Given n = 10, d = 100 and S10 = 8500.

  Find the value of a1.

 

  Find the value of a10.

 

 

Find the sum of each arithmetic series. 

57. 

 

SOLUTION:  There are 16 – 1 + 1 or 16 terms, so n = 16. 

  Find the sum.  

 

Therefore, .

 

58. 

 

SOLUTION:  There are 13 – 4 + 1 or 10 terms, so n = 10. 

  Find the sum.  

 

Therefore, .

 

59. 

 

SOLUTION:  There are 16 – 5 + 1 or 12 terms, so n = 12. 

  Find the sum.  

 

Therefore, .

 

60. 

 

SOLUTION:  There are 12 – 0 + 1 or 13 terms, so n = 13. 

  Find the sum.  

 

Therefore, .

 

61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end ofthe first month and $25 more each additional month for 12 months. How much does she pay in total after the 12 months?  

SOLUTION:  

Given a1 = 50, d = 25 and n = 12.

  Find the sum.  

  She pays $2250.  

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?  

SOLUTION:  

Given a1 = 16, n = 10.

  The common difference d is 32.   Find the sum.  

  The object will fall 1600 ft in 10 seconds. 

Use the given information to write an equation that represents the nth term in each arithmetic sequence.  

63. The 100th term of the sequence is 245. The common difference is 13. 

SOLUTION:  

Given a100 = 245, d = 13 and n = 100.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

64. The eleventh term of the sequence is 78. The common difference is –9. 

SOLUTION:  

Given a11 = 78, d = –9 and n = 11.

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

65. The sixth term of the sequence is –34. The 23rd term is 119.  

SOLUTION:  

Given a6 = –34 and a23 = 119.

  Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.   Find the common difference of the series with a1 = –34 and a18 = 119.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

66. The 25th term of the sequence is 121. The 80th term is 506.  

SOLUTION:  

Given a25 = 121 and a80 = 506.

  Therefore, there are (80 – 25 + 1) or 56 terms between 121 and 506.   Find the common difference of the series with a1 = 121 and a56 = 506.

 

  Find the value of a1.

 

  Substitute the values of a1 and d to find the nth term.

 

 

67. CCSS MODELING  The rectangular tables in a reception hall are often placed end-to-end to form one long table.The diagrams below show the number of people who can sit at each of the table arrangements.  

  a. Make drawings to find the next three numbers as tables are added one at a time to the arrangement.   b. Write an equation representing the nth number in this pattern.   c. Is it possible to have seating for exactly 100 people with such an arrangement? Explain.  

SOLUTION:  a. For each increase in the number of table, the number of people who can sit is increased by 4. That is, the common difference is 4. Therefore, the next three numbers are (10 + 4) or 14, (14 + 4) or 18 and (18 + 4) or 22.

  b. Substitute a1 = 6 and d = 4 in .

  c. No; there is no whole number n for which .  

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?  

SOLUTION:  

Given a1 = 200, d = 24 and an = 512.

  Substitute the values of a1, an and d and solve for n.

 

  14th term in the sequence is 512.   Therefore, she completed (14 – 1) or 13 units.  

69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how many years will he have a salary of $100,000 per year?  

SOLUTION:  

Given a1 = 28000, d = 4000 and an = 100000.

  Substitute the values of a1, an and d and solve for n.

 

  So he will have a salary of $100,000 per year after the 19th year. 

70. SPORTS While training for cross country, Silvia plans to run 3 miles per day for the first week, and then increase the distance by a half mile each of the following weeks.   a. Write an equation to represent the nth term of the sequence.   b. If the pattern continues, during which week will she be running 10 miles per day?   c. Is it reasonable for this pattern to continue indefinitely? Explain.  

SOLUTION:  a. Given a1 = 3 and d = 0.5.

  Find the nth term.  

  b. Substitute 10 for an in  and solve for n.

 

  During 15th week, she will be running 10 miles per day.   c. Sample answer: No; eventually the number of miles per day will become unrealistic. 

71. MUTIPLE REPRESENTATIONS Consider

  a. TABULAR Make a table of the partial sums of the series for 1 ≤ k  ≤ 10.   b. GRAPHICAL Graph (k , partial sum).  

c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.

  d. VERBAL What do you notice about the two graphs?   e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum of arithmetic series?   f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x

2 + 8x.

 

SOLUTION:  a.

  b.

  c.

  d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.   e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that shares the same range.  

f.

Find the value of x.

 

72. 

 

SOLUTION:  There are x – 3 + 1 or x – 2 terms, so n = x – 2.  

  Find the sum.  

  Equate the sum with the given value and solve for x. 

 

  The value of x should be positive. Therefore, x = 18. 

73. 

 

SOLUTION:  There are x – 5 + 1 or x – 4 terms, so n = x – 4.  

  Find the sum.  

  Equate the sum with the given value and solve for x.

   

  The value of x should be positive. Therefore, x = 16. 

74. CCSS CRITIQUE  Eric and Juana are determining the formula for the nth term for the sequence –11, –2, 7, 16, … . Is either of them correct? Explain your reasoning.  

 

SOLUTION:  Sample answer: Eric; Juana missed the step of multiplying d by n – 1. 

75. REASONING If a is the third term in an arithmetic sequence, b is the fifth term, and c is the eleventh term, express c in terms of a and b.  

SOLUTION:  

Given a3 = a, a5 = b and a11 = c.

  Find the common difference.  

  Find the value of a1.

 

  Find the value of c in terms of a and b. 

 

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?  

SOLUTION:  

The three arithmetic means between a and b are .

 

The average of the arithmetic means is .

  Therefore, . The term b can be written as a + 4d.  

The average of a and b is .

  We know that .   Therefore, the average of a and b is 16.  

77. CHALLENGE Find Sn for (x + y) + (x + 2y) + (x + 3y) + … .

 

SOLUTION:  (x + 2y) – (x + y) = y (x + 3y) – (x + 2y) = y   The common difference is y . 

  Find the sum.  

 

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. 

SOLUTION:  Sample answer: 9 + 18 + 27 + … + 72 

79. WRITING IN MATH Compare and contrast arithmetic sequences and series.  

SOLUTION:  Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence.  

80. PROOF Prove the formula for the nth term of an arithmetic sequence. 

SOLUTION:  Sample answer: Let an = the nth term of the sequence and d = the common difference

a2 = a1 + d Definition of the second term of an arithmetic sequence

a3 = a2 + d Definition of the third term of an arithmetic sequence

a3 = (a1 + d ) + d Substitution

a3 = a1 + 2d Associative Property of Addition

a3 = a1 + (3 – 1)d      3 – 1 = 2

an = a1 + (n – 1)d      n = 3

 

81. PROOF Derive a sum formula that does not include a1.

 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

  an

– (n – 1)d = a1    Subtract (n – 1)d from both sides.

 

 Substitution

 

 Simplify.

 

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula. 

SOLUTION:  

General sum formula

  an = a1 + (n – 1)d   Formula for nth term

 

 Substitution

 

 Simplify.

 

83. SAT/ACT  The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest angle is 36°, what is the measure of the largest angle?  

  A 54°   B 75°   C 84°   D 90°   E 97°  

SOLUTION:  The sum of the interior angle of a triangle is 180°. Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d. 36 + 36 + d + 36 + 2d = 180

d = 24

Therefore, the largest angle is 36° + 48° = 84°.   Option C is the correct answer.

84. The area of a triangle is  and the height is q + 4. Which expression best describes the triangle’s length?

  F  (q + 1)   G (q + 2)     H (q – 3)   J  (q – 4)  

SOLUTION:  

The area of the triangle is .

 

The length of the triangle is

  Option J is the correct answer.  

85. The expression  is equivalent to 

A

 

B

 

C

 

D

 

SOLUTION:  

  Option A is the correct answer. 

86. SHORT RESPONSE Trevor can type a 200-word essay in 6 hours. Minya can type the same essay in hours.

If they work together, how many hours will it take them to type the essay?  

SOLUTION:  

Trevor can type  words in an hour.

 

Minya can type  words in an hour.

 

They type words together in an hour.

 

 

They will type the essay in  hours if they work together.

 

Determine whether each sequence is arithmetic. Write yes or no.

  87. –6, 4, 14, 24, …

 

SOLUTION:  Since there is a common difference between the consecutive terms, this is an arithmetic sequence. 

88. 

 

SOLUTION:  Since there is a common difference between the consecutive terms, this is an arithmetic sequence. 

89. 10, 8, 5, 1, ...  

SOLUTION:  Since there is no common difference between the consecutive terms, this is not an arithmetic sequence. 

Solve each system of inequalities by graphing. 

90. 

 

SOLUTION:  

 

91. 

 

SOLUTION:  

 

92. 

 

SOLUTION:  

 

93. PHYSICS The distance a spring stretches is related to the mass attached to the spring. This is represented by d =

km, where d is the distance, m is the mass, and k is the spring constant. When two springs with spring constants k1

and k2 are attached in a series, the resulting spring constant k is found by the equation

  a. If one spring with constant of 12 centimeters per gram is attached in a series with another spring with constant of8 centimeters per gram, find the resultant spring constant.   b. If a 5-gram object is hung from the series of springs, how far will the springs stretch? Is this answer reasonable in this context?  

 

SOLUTION:  a. Given k1 = 12 and k2 = 8.

  Substitute the values and evaluate the value of k .

  b. Substitute 4.8 and 5 for k and m respectively in the equation d = km.  

  The answer is reasonable. The object would stretch the first spring 60 cm and would stretch the second spring 40 cm. The object would have to stretch the combined springs less than it would stretch either of the springs individually.  

Graph each function. State the domain and range. 

94. 

 

SOLUTION:  Graph the function.  

  The function is defined for all values of x.   Therefore, the domain is D = {all real numbers}.   The value of the f (x) tends to 0 as x tends to –∞.   The value of the f (x) tends to ∞ as x tends to ∞.   Therefore, the range of the function is R ={f (x) | f (x) > 0}. 

95. 

 

SOLUTION:  Graph the function.  

  The function is defined for all values of x.   Therefore, the domain is D = {all real numbers}.   The value of the f (x) tends to 3 as x tends to –∞.   The value of the f (x) tends to ∞ as x tends to ∞.   Therefore, the range of the function is R ={f (x) | f (x) > 3}. 

96. 

 

SOLUTION:  Graph the function.  

  The function is defined for all values of x.   Therefore, the domain is D = {all real numbers}.   The value of the f (x) tends to ∞ as x tends to –∞.   The value of the f (x) tends to –1 as x tends to ∞.   Therefore, the range of the function is R ={f (x) | f (x) > –1}. 

Solve each equation. Round to the nearest ten-thousandth. 

97. 5x = 52

 

SOLUTION:  

 

98. 43p = 10

 

SOLUTION:  

 

99. 3n + 2

= 14.5

 

SOLUTION:  

 

100. 16d – 4

= 33 – d

 

SOLUTION:  

 

eSolutions Manual - Powered by Cognero Page 49

10-2 Arithmetic Sequences and Series

  H (q – 3)   J  (q – 4)  

SOLUTION:  

The area of the triangle is .

 

The length of the triangle is

  Option J is the correct answer.  

85. The expression  is equivalent to 

A

 

B

 

C

 

D

 

SOLUTION:  

  Option A is the correct answer. 

86. SHORT RESPONSE Trevor can type a 200-word essay in 6 hours. Minya can type the same essay in hours.

If they work together, how many hours will it take them to type the essay?  

SOLUTION:  

Trevor can type  words in an hour.

 

Minya can type  words in an hour.

 

They type words together in an hour.

 

 

They will type the essay in  hours if they work together.

 

Determine whether each sequence is arithmetic. Write yes or no.

  87. –6, 4, 14, 24, …

 

SOLUTION:  Since there is a common difference between the consecutive terms, this is an arithmetic sequence. 

88. 

 

SOLUTION:  Since there is a common difference between the consecutive terms, this is an arithmetic sequence. 

89. 10, 8, 5, 1, ...  

SOLUTION:  Since there is no common difference between the consecutive terms, this is not an arithmetic sequence. 

Solve each system of inequalities by graphing. 

90. 

 

SOLUTION:  

 

91. 

 

SOLUTION:  

 

92. 

 

SOLUTION:  

 

93. PHYSICS The distance a spring stretches is related to the mass attached to the spring. This is represented by d =

km, where d is the distance, m is the mass, and k is the spring constant. When two springs with spring constants k1

and k2 are attached in a series, the resulting spring constant k is found by the equation

  a. If one spring with constant of 12 centimeters per gram is attached in a series with another spring with constant of8 centimeters per gram, find the resultant spring constant.   b. If a 5-gram object is hung from the series of springs, how far will the springs stretch? Is this answer reasonable in this context?  

 

SOLUTION:  a. Given k1 = 12 and k2 = 8.

  Substitute the values and evaluate the value of k .

  b. Substitute 4.8 and 5 for k and m respectively in the equation d = km.  

  The answer is reasonable. The object would stretch the first spring 60 cm and would stretch the second spring 40 cm. The object would have to stretch the combined springs less than it would stretch either of the springs individually.  

Graph each function. State the domain and range. 

94. 

 

SOLUTION:  Graph the function.  

  The function is defined for all values of x.   Therefore, the domain is D = {all real numbers}.   The value of the f (x) tends to 0 as x tends to –∞.   The value of the f (x) tends to ∞ as x tends to ∞.   Therefore, the range of the function is R ={f (x) | f (x) > 0}. 

95. 

 

SOLUTION:  Graph the function.  

  The function is defined for all values of x.   Therefore, the domain is D = {all real numbers}.   The value of the f (x) tends to 3 as x tends to –∞.   The value of the f (x) tends to ∞ as x tends to ∞.   Therefore, the range of the function is R ={f (x) | f (x) > 3}. 

96. 

 

SOLUTION:  Graph the function.  

  The function is defined for all values of x.   Therefore, the domain is D = {all real numbers}.   The value of the f (x) tends to ∞ as x tends to –∞.   The value of the f (x) tends to –1 as x tends to ∞.   Therefore, the range of the function is R ={f (x) | f (x) > –1}. 

Solve each equation. Round to the nearest ten-thousandth. 

97. 5x = 52

 

SOLUTION:  

 

98. 43p = 10

 

SOLUTION:  

 

99. 3n + 2

= 14.5

 

SOLUTION:  

 

100. 16d – 4

= 33 – d

 

SOLUTION:  

 

eSolutions Manual - Powered by Cognero Page 50

10-2 Arithmetic Sequences and Series

  H (q – 3)   J  (q – 4)  

SOLUTION:  

The area of the triangle is .

 

The length of the triangle is

  Option J is the correct answer.  

85. The expression  is equivalent to 

A

 

B

 

C

 

D

 

SOLUTION:  

  Option A is the correct answer. 

86. SHORT RESPONSE Trevor can type a 200-word essay in 6 hours. Minya can type the same essay in hours.

If they work together, how many hours will it take them to type the essay?  

SOLUTION:  

Trevor can type  words in an hour.

 

Minya can type  words in an hour.

 

They type words together in an hour.

 

 

They will type the essay in  hours if they work together.

 

Determine whether each sequence is arithmetic. Write yes or no.

  87. –6, 4, 14, 24, …

 

SOLUTION:  Since there is a common difference between the consecutive terms, this is an arithmetic sequence. 

88. 

 

SOLUTION:  Since there is a common difference between the consecutive terms, this is an arithmetic sequence. 

89. 10, 8, 5, 1, ...  

SOLUTION:  Since there is no common difference between the consecutive terms, this is not an arithmetic sequence. 

Solve each system of inequalities by graphing. 

90. 

 

SOLUTION:  

 

91. 

 

SOLUTION:  

 

92. 

 

SOLUTION:  

 

93. PHYSICS The distance a spring stretches is related to the mass attached to the spring. This is represented by d =

km, where d is the distance, m is the mass, and k is the spring constant. When two springs with spring constants k1

and k2 are attached in a series, the resulting spring constant k is found by the equation

  a. If one spring with constant of 12 centimeters per gram is attached in a series with another spring with constant of8 centimeters per gram, find the resultant spring constant.   b. If a 5-gram object is hung from the series of springs, how far will the springs stretch? Is this answer reasonable in this context?  

 

SOLUTION:  a. Given k1 = 12 and k2 = 8.

  Substitute the values and evaluate the value of k .

  b. Substitute 4.8 and 5 for k and m respectively in the equation d = km.  

  The answer is reasonable. The object would stretch the first spring 60 cm and would stretch the second spring 40 cm. The object would have to stretch the combined springs less than it would stretch either of the springs individually.  

Graph each function. State the domain and range. 

94. 

 

SOLUTION:  Graph the function.  

  The function is defined for all values of x.   Therefore, the domain is D = {all real numbers}.   The value of the f (x) tends to 0 as x tends to –∞.   The value of the f (x) tends to ∞ as x tends to ∞.   Therefore, the range of the function is R ={f (x) | f (x) > 0}. 

95. 

 

SOLUTION:  Graph the function.  

  The function is defined for all values of x.   Therefore, the domain is D = {all real numbers}.   The value of the f (x) tends to 3 as x tends to –∞.   The value of the f (x) tends to ∞ as x tends to ∞.   Therefore, the range of the function is R ={f (x) | f (x) > 3}. 

96. 

 

SOLUTION:  Graph the function.  

  The function is defined for all values of x.   Therefore, the domain is D = {all real numbers}.   The value of the f (x) tends to ∞ as x tends to –∞.   The value of the f (x) tends to –1 as x tends to ∞.   Therefore, the range of the function is R ={f (x) | f (x) > –1}. 

Solve each equation. Round to the nearest ten-thousandth. 

97. 5x = 52

 

SOLUTION:  

 

98. 43p = 10

 

SOLUTION:  

 

99. 3n + 2

= 14.5

 

SOLUTION:  

 

100. 16d – 4

= 33 – d

 

SOLUTION:  

 

eSolutions Manual - Powered by Cognero Page 51

10-2 Arithmetic Sequences and Series

  H (q – 3)   J  (q – 4)  

SOLUTION:  

The area of the triangle is .

 

The length of the triangle is

  Option J is the correct answer.  

85. The expression  is equivalent to 

A

 

B

 

C

 

D

 

SOLUTION:  

  Option A is the correct answer. 

86. SHORT RESPONSE Trevor can type a 200-word essay in 6 hours. Minya can type the same essay in hours.

If they work together, how many hours will it take them to type the essay?  

SOLUTION:  

Trevor can type  words in an hour.

 

Minya can type  words in an hour.

 

They type words together in an hour.

 

 

They will type the essay in  hours if they work together.

 

Determine whether each sequence is arithmetic. Write yes or no.

  87. –6, 4, 14, 24, …

 

SOLUTION:  Since there is a common difference between the consecutive terms, this is an arithmetic sequence. 

88. 

 

SOLUTION:  Since there is a common difference between the consecutive terms, this is an arithmetic sequence. 

89. 10, 8, 5, 1, ...  

SOLUTION:  Since there is no common difference between the consecutive terms, this is not an arithmetic sequence. 

Solve each system of inequalities by graphing. 

90. 

 

SOLUTION:  

 

91. 

 

SOLUTION:  

 

92. 

 

SOLUTION:  

 

93. PHYSICS The distance a spring stretches is related to the mass attached to the spring. This is represented by d =

km, where d is the distance, m is the mass, and k is the spring constant. When two springs with spring constants k1

and k2 are attached in a series, the resulting spring constant k is found by the equation

  a. If one spring with constant of 12 centimeters per gram is attached in a series with another spring with constant of8 centimeters per gram, find the resultant spring constant.   b. If a 5-gram object is hung from the series of springs, how far will the springs stretch? Is this answer reasonable in this context?  

 

SOLUTION:  a. Given k1 = 12 and k2 = 8.

  Substitute the values and evaluate the value of k .

  b. Substitute 4.8 and 5 for k and m respectively in the equation d = km.  

  The answer is reasonable. The object would stretch the first spring 60 cm and would stretch the second spring 40 cm. The object would have to stretch the combined springs less than it would stretch either of the springs individually.  

Graph each function. State the domain and range. 

94. 

 

SOLUTION:  Graph the function.  

  The function is defined for all values of x.   Therefore, the domain is D = {all real numbers}.   The value of the f (x) tends to 0 as x tends to –∞.   The value of the f (x) tends to ∞ as x tends to ∞.   Therefore, the range of the function is R ={f (x) | f (x) > 0}. 

95. 

 

SOLUTION:  Graph the function.  

  The function is defined for all values of x.   Therefore, the domain is D = {all real numbers}.   The value of the f (x) tends to 3 as x tends to –∞.   The value of the f (x) tends to ∞ as x tends to ∞.   Therefore, the range of the function is R ={f (x) | f (x) > 3}. 

96. 

 

SOLUTION:  Graph the function.  

  The function is defined for all values of x.   Therefore, the domain is D = {all real numbers}.   The value of the f (x) tends to ∞ as x tends to –∞.   The value of the f (x) tends to –1 as x tends to ∞.   Therefore, the range of the function is R ={f (x) | f (x) > –1}. 

Solve each equation. Round to the nearest ten-thousandth. 

97. 5x = 52

 

SOLUTION:  

 

98. 43p = 10

 

SOLUTION:  

 

99. 3n + 2

= 14.5

 

SOLUTION:  

 

100. 16d – 4

= 33 – d

 

SOLUTION:  

 

eSolutions Manual - Powered by Cognero Page 52

10-2 Arithmetic Sequences and Series

  H (q – 3)   J  (q – 4)  

SOLUTION:  

The area of the triangle is .

 

The length of the triangle is

  Option J is the correct answer.  

85. The expression  is equivalent to 

A

 

B

 

C

 

D

 

SOLUTION:  

  Option A is the correct answer. 

86. SHORT RESPONSE Trevor can type a 200-word essay in 6 hours. Minya can type the same essay in hours.

If they work together, how many hours will it take them to type the essay?  

SOLUTION:  

Trevor can type  words in an hour.

 

Minya can type  words in an hour.

 

They type words together in an hour.

 

 

They will type the essay in  hours if they work together.

 

Determine whether each sequence is arithmetic. Write yes or no.

  87. –6, 4, 14, 24, …

 

SOLUTION:  Since there is a common difference between the consecutive terms, this is an arithmetic sequence. 

88. 

 

SOLUTION:  Since there is a common difference between the consecutive terms, this is an arithmetic sequence. 

89. 10, 8, 5, 1, ...  

SOLUTION:  Since there is no common difference between the consecutive terms, this is not an arithmetic sequence. 

Solve each system of inequalities by graphing. 

90. 

 

SOLUTION:  

 

91. 

 

SOLUTION:  

 

92. 

 

SOLUTION:  

 

93. PHYSICS The distance a spring stretches is related to the mass attached to the spring. This is represented by d =

km, where d is the distance, m is the mass, and k is the spring constant. When two springs with spring constants k1

and k2 are attached in a series, the resulting spring constant k is found by the equation

  a. If one spring with constant of 12 centimeters per gram is attached in a series with another spring with constant of8 centimeters per gram, find the resultant spring constant.   b. If a 5-gram object is hung from the series of springs, how far will the springs stretch? Is this answer reasonable in this context?  

 

SOLUTION:  a. Given k1 = 12 and k2 = 8.

  Substitute the values and evaluate the value of k .

  b. Substitute 4.8 and 5 for k and m respectively in the equation d = km.  

  The answer is reasonable. The object would stretch the first spring 60 cm and would stretch the second spring 40 cm. The object would have to stretch the combined springs less than it would stretch either of the springs individually.  

Graph each function. State the domain and range. 

94. 

 

SOLUTION:  Graph the function.  

  The function is defined for all values of x.   Therefore, the domain is D = {all real numbers}.   The value of the f (x) tends to 0 as x tends to –∞.   The value of the f (x) tends to ∞ as x tends to ∞.   Therefore, the range of the function is R ={f (x) | f (x) > 0}. 

95. 

 

SOLUTION:  Graph the function.  

  The function is defined for all values of x.   Therefore, the domain is D = {all real numbers}.   The value of the f (x) tends to 3 as x tends to –∞.   The value of the f (x) tends to ∞ as x tends to ∞.   Therefore, the range of the function is R ={f (x) | f (x) > 3}. 

96. 

 

SOLUTION:  Graph the function.  

  The function is defined for all values of x.   Therefore, the domain is D = {all real numbers}.   The value of the f (x) tends to ∞ as x tends to –∞.   The value of the f (x) tends to –1 as x tends to ∞.   Therefore, the range of the function is R ={f (x) | f (x) > –1}. 

Solve each equation. Round to the nearest ten-thousandth. 

97. 5x = 52

 

SOLUTION:  

 

98. 43p = 10

 

SOLUTION:  

 

99. 3n + 2

= 14.5

 

SOLUTION:  

 

100. 16d – 4

= 33 – d

 

SOLUTION:  

 

eSolutions Manual - Powered by Cognero Page 53

10-2 Arithmetic Sequences and Series

  H (q – 3)   J  (q – 4)  

SOLUTION:  

The area of the triangle is .

 

The length of the triangle is

  Option J is the correct answer.  

85. The expression  is equivalent to 

A

 

B

 

C

 

D

 

SOLUTION:  

  Option A is the correct answer. 

86. SHORT RESPONSE Trevor can type a 200-word essay in 6 hours. Minya can type the same essay in hours.

If they work together, how many hours will it take them to type the essay?  

SOLUTION:  

Trevor can type  words in an hour.

 

Minya can type  words in an hour.

 

They type words together in an hour.

 

 

They will type the essay in  hours if they work together.

 

Determine whether each sequence is arithmetic. Write yes or no.

  87. –6, 4, 14, 24, …

 

SOLUTION:  Since there is a common difference between the consecutive terms, this is an arithmetic sequence. 

88. 

 

SOLUTION:  Since there is a common difference between the consecutive terms, this is an arithmetic sequence. 

89. 10, 8, 5, 1, ...  

SOLUTION:  Since there is no common difference between the consecutive terms, this is not an arithmetic sequence. 

Solve each system of inequalities by graphing. 

90. 

 

SOLUTION:  

 

91. 

 

SOLUTION:  

 

92. 

 

SOLUTION:  

 

93. PHYSICS The distance a spring stretches is related to the mass attached to the spring. This is represented by d =

km, where d is the distance, m is the mass, and k is the spring constant. When two springs with spring constants k1

and k2 are attached in a series, the resulting spring constant k is found by the equation

  a. If one spring with constant of 12 centimeters per gram is attached in a series with another spring with constant of8 centimeters per gram, find the resultant spring constant.   b. If a 5-gram object is hung from the series of springs, how far will the springs stretch? Is this answer reasonable in this context?  

 

SOLUTION:  a. Given k1 = 12 and k2 = 8.

  Substitute the values and evaluate the value of k .

  b. Substitute 4.8 and 5 for k and m respectively in the equation d = km.  

  The answer is reasonable. The object would stretch the first spring 60 cm and would stretch the second spring 40 cm. The object would have to stretch the combined springs less than it would stretch either of the springs individually.  

Graph each function. State the domain and range. 

94. 

 

SOLUTION:  Graph the function.  

  The function is defined for all values of x.   Therefore, the domain is D = {all real numbers}.   The value of the f (x) tends to 0 as x tends to –∞.   The value of the f (x) tends to ∞ as x tends to ∞.   Therefore, the range of the function is R ={f (x) | f (x) > 0}. 

95. 

 

SOLUTION:  Graph the function.  

  The function is defined for all values of x.   Therefore, the domain is D = {all real numbers}.   The value of the f (x) tends to 3 as x tends to –∞.   The value of the f (x) tends to ∞ as x tends to ∞.   Therefore, the range of the function is R ={f (x) | f (x) > 3}. 

96. 

 

SOLUTION:  Graph the function.  

  The function is defined for all values of x.   Therefore, the domain is D = {all real numbers}.   The value of the f (x) tends to ∞ as x tends to –∞.   The value of the f (x) tends to –1 as x tends to ∞.   Therefore, the range of the function is R ={f (x) | f (x) > –1}. 

Solve each equation. Round to the nearest ten-thousandth. 

97. 5x = 52

 

SOLUTION:  

 

98. 43p = 10

 

SOLUTION:  

 

99. 3n + 2

= 14.5

 

SOLUTION:  

 

100. 16d – 4

= 33 – d

 

SOLUTION:  

 

eSolutions Manual - Powered by Cognero Page 54

10-2 Arithmetic Sequences and Series

  H (q – 3)   J  (q – 4)  

SOLUTION:  

The area of the triangle is .

 

The length of the triangle is

  Option J is the correct answer.  

85. The expression  is equivalent to 

A

 

B

 

C

 

D

 

SOLUTION:  

  Option A is the correct answer. 

86. SHORT RESPONSE Trevor can type a 200-word essay in 6 hours. Minya can type the same essay in hours.

If they work together, how many hours will it take them to type the essay?  

SOLUTION:  

Trevor can type  words in an hour.

 

Minya can type  words in an hour.

 

They type words together in an hour.

 

 

They will type the essay in  hours if they work together.

 

Determine whether each sequence is arithmetic. Write yes or no.

  87. –6, 4, 14, 24, …

 

SOLUTION:  Since there is a common difference between the consecutive terms, this is an arithmetic sequence. 

88. 

 

SOLUTION:  Since there is a common difference between the consecutive terms, this is an arithmetic sequence. 

89. 10, 8, 5, 1, ...  

SOLUTION:  Since there is no common difference between the consecutive terms, this is not an arithmetic sequence. 

Solve each system of inequalities by graphing. 

90. 

 

SOLUTION:  

 

91. 

 

SOLUTION:  

 

92. 

 

SOLUTION:  

 

93. PHYSICS The distance a spring stretches is related to the mass attached to the spring. This is represented by d =

km, where d is the distance, m is the mass, and k is the spring constant. When two springs with spring constants k1

and k2 are attached in a series, the resulting spring constant k is found by the equation

  a. If one spring with constant of 12 centimeters per gram is attached in a series with another spring with constant of8 centimeters per gram, find the resultant spring constant.   b. If a 5-gram object is hung from the series of springs, how far will the springs stretch? Is this answer reasonable in this context?  

 

SOLUTION:  a. Given k1 = 12 and k2 = 8.

  Substitute the values and evaluate the value of k .

  b. Substitute 4.8 and 5 for k and m respectively in the equation d = km.  

  The answer is reasonable. The object would stretch the first spring 60 cm and would stretch the second spring 40 cm. The object would have to stretch the combined springs less than it would stretch either of the springs individually.  

Graph each function. State the domain and range. 

94. 

 

SOLUTION:  Graph the function.  

  The function is defined for all values of x.   Therefore, the domain is D = {all real numbers}.   The value of the f (x) tends to 0 as x tends to –∞.   The value of the f (x) tends to ∞ as x tends to ∞.   Therefore, the range of the function is R ={f (x) | f (x) > 0}. 

95. 

 

SOLUTION:  Graph the function.  

  The function is defined for all values of x.   Therefore, the domain is D = {all real numbers}.   The value of the f (x) tends to 3 as x tends to –∞.   The value of the f (x) tends to ∞ as x tends to ∞.   Therefore, the range of the function is R ={f (x) | f (x) > 3}. 

96. 

 

SOLUTION:  Graph the function.  

  The function is defined for all values of x.   Therefore, the domain is D = {all real numbers}.   The value of the f (x) tends to ∞ as x tends to –∞.   The value of the f (x) tends to –1 as x tends to ∞.   Therefore, the range of the function is R ={f (x) | f (x) > –1}. 

Solve each equation. Round to the nearest ten-thousandth. 

97. 5x = 52

 

SOLUTION:  

 

98. 43p = 10

 

SOLUTION:  

 

99. 3n + 2

= 14.5

 

SOLUTION:  

 

100. 16d – 4

= 33 – d

 

SOLUTION:  

 

eSolutions Manual - Powered by Cognero Page 55

10-2 Arithmetic Sequences and Series

  H (q – 3)   J  (q – 4)  

SOLUTION:  

The area of the triangle is .

 

The length of the triangle is

  Option J is the correct answer.  

85. The expression  is equivalent to 

A

 

B

 

C

 

D

 

SOLUTION:  

  Option A is the correct answer. 

86. SHORT RESPONSE Trevor can type a 200-word essay in 6 hours. Minya can type the same essay in hours.

If they work together, how many hours will it take them to type the essay?  

SOLUTION:  

Trevor can type  words in an hour.

 

Minya can type  words in an hour.

 

They type words together in an hour.

 

 

They will type the essay in  hours if they work together.

 

Determine whether each sequence is arithmetic. Write yes or no.

  87. –6, 4, 14, 24, …

 

SOLUTION:  Since there is a common difference between the consecutive terms, this is an arithmetic sequence. 

88. 

 

SOLUTION:  Since there is a common difference between the consecutive terms, this is an arithmetic sequence. 

89. 10, 8, 5, 1, ...  

SOLUTION:  Since there is no common difference between the consecutive terms, this is not an arithmetic sequence. 

Solve each system of inequalities by graphing. 

90. 

 

SOLUTION:  

 

91. 

 

SOLUTION:  

 

92. 

 

SOLUTION:  

 

93. PHYSICS The distance a spring stretches is related to the mass attached to the spring. This is represented by d =

km, where d is the distance, m is the mass, and k is the spring constant. When two springs with spring constants k1

and k2 are attached in a series, the resulting spring constant k is found by the equation

  a. If one spring with constant of 12 centimeters per gram is attached in a series with another spring with constant of8 centimeters per gram, find the resultant spring constant.   b. If a 5-gram object is hung from the series of springs, how far will the springs stretch? Is this answer reasonable in this context?  

 

SOLUTION:  a. Given k1 = 12 and k2 = 8.

  Substitute the values and evaluate the value of k .

  b. Substitute 4.8 and 5 for k and m respectively in the equation d = km.  

  The answer is reasonable. The object would stretch the first spring 60 cm and would stretch the second spring 40 cm. The object would have to stretch the combined springs less than it would stretch either of the springs individually.  

Graph each function. State the domain and range. 

94. 

 

SOLUTION:  Graph the function.  

  The function is defined for all values of x.   Therefore, the domain is D = {all real numbers}.   The value of the f (x) tends to 0 as x tends to –∞.   The value of the f (x) tends to ∞ as x tends to ∞.   Therefore, the range of the function is R ={f (x) | f (x) > 0}. 

95. 

 

SOLUTION:  Graph the function.  

  The function is defined for all values of x.   Therefore, the domain is D = {all real numbers}.   The value of the f (x) tends to 3 as x tends to –∞.   The value of the f (x) tends to ∞ as x tends to ∞.   Therefore, the range of the function is R ={f (x) | f (x) > 3}. 

96. 

 

SOLUTION:  Graph the function.  

  The function is defined for all values of x.   Therefore, the domain is D = {all real numbers}.   The value of the f (x) tends to ∞ as x tends to –∞.   The value of the f (x) tends to –1 as x tends to ∞.   Therefore, the range of the function is R ={f (x) | f (x) > –1}. 

Solve each equation. Round to the nearest ten-thousandth. 

97. 5x = 52

 

SOLUTION:  

 

98. 43p = 10

 

SOLUTION:  

 

99. 3n + 2

= 14.5

 

SOLUTION:  

 

100. 16d – 4

= 33 – d

 

SOLUTION:  

 

eSolutions Manual - Powered by Cognero Page 56

10-2 Arithmetic Sequences and Series

  H (q – 3)   J  (q – 4)  

SOLUTION:  

The area of the triangle is .

 

The length of the triangle is

  Option J is the correct answer.  

85. The expression  is equivalent to 

A

 

B

 

C

 

D

 

SOLUTION:  

  Option A is the correct answer. 

86. SHORT RESPONSE Trevor can type a 200-word essay in 6 hours. Minya can type the same essay in hours.

If they work together, how many hours will it take them to type the essay?  

SOLUTION:  

Trevor can type  words in an hour.

 

Minya can type  words in an hour.

 

They type words together in an hour.

 

 

They will type the essay in  hours if they work together.

 

Determine whether each sequence is arithmetic. Write yes or no.

  87. –6, 4, 14, 24, …

 

SOLUTION:  Since there is a common difference between the consecutive terms, this is an arithmetic sequence. 

88. 

 

SOLUTION:  Since there is a common difference between the consecutive terms, this is an arithmetic sequence. 

89. 10, 8, 5, 1, ...  

SOLUTION:  Since there is no common difference between the consecutive terms, this is not an arithmetic sequence. 

Solve each system of inequalities by graphing. 

90. 

 

SOLUTION:  

 

91. 

 

SOLUTION:  

 

92. 

 

SOLUTION:  

 

93. PHYSICS The distance a spring stretches is related to the mass attached to the spring. This is represented by d =

km, where d is the distance, m is the mass, and k is the spring constant. When two springs with spring constants k1

and k2 are attached in a series, the resulting spring constant k is found by the equation

  a. If one spring with constant of 12 centimeters per gram is attached in a series with another spring with constant of8 centimeters per gram, find the resultant spring constant.   b. If a 5-gram object is hung from the series of springs, how far will the springs stretch? Is this answer reasonable in this context?  

 

SOLUTION:  a. Given k1 = 12 and k2 = 8.

  Substitute the values and evaluate the value of k .

  b. Substitute 4.8 and 5 for k and m respectively in the equation d = km.  

  The answer is reasonable. The object would stretch the first spring 60 cm and would stretch the second spring 40 cm. The object would have to stretch the combined springs less than it would stretch either of the springs individually.  

Graph each function. State the domain and range. 

94. 

 

SOLUTION:  Graph the function.  

  The function is defined for all values of x.   Therefore, the domain is D = {all real numbers}.   The value of the f (x) tends to 0 as x tends to –∞.   The value of the f (x) tends to ∞ as x tends to ∞.   Therefore, the range of the function is R ={f (x) | f (x) > 0}. 

95. 

 

SOLUTION:  Graph the function.  

  The function is defined for all values of x.   Therefore, the domain is D = {all real numbers}.   The value of the f (x) tends to 3 as x tends to –∞.   The value of the f (x) tends to ∞ as x tends to ∞.   Therefore, the range of the function is R ={f (x) | f (x) > 3}. 

96. 

 

SOLUTION:  Graph the function.  

  The function is defined for all values of x.   Therefore, the domain is D = {all real numbers}.   The value of the f (x) tends to ∞ as x tends to –∞.   The value of the f (x) tends to –1 as x tends to ∞.   Therefore, the range of the function is R ={f (x) | f (x) > –1}. 

Solve each equation. Round to the nearest ten-thousandth. 

97. 5x = 52

 

SOLUTION:  

 

98. 43p = 10

 

SOLUTION:  

 

99. 3n + 2

= 14.5

 

SOLUTION:  

 

100. 16d – 4

= 33 – d

 

SOLUTION:  

 

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10-2 Arithmetic Sequences and Series