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UNIVERSITI TEKNOLOGI MARA

MARIATHY KARIM

UiTM SARAWAK (SAMARAHAN)

FSKM(MATHEMATICS)

A set of numbers is arranged in a definite order where there exists a first term, a second term, a third term and so on. Whereby each term is obtained by following a specific formula or a special rule. This set of numbers is called a sequence and also known as progression.

There are two types of sequences : Arithmetic and Geometric Sequence.

Sequence: T1, T2, T3, , Tn.

T1 is the first term. Also defined as a.

Arithmetic Sequence

Definition

Sequence of numbers whereby each term(except the first term) is obtained by adding a constant or a fixed

number known as common difference (d) to its immediate predecessor.

Common Difference

The common difference can be either positive or negative value.

Common Difference , d = Tn Tn-1

Examples: d = T2 T1 or d = T3 T2

The nth term

Tn = a + (n 1)d

Sum of the first n terms

dnanSn )1(22

To determine common difference in an arithmetic sequence; any term can be choose and then subtract the value of the term with its immediate predecessor.

Example:

1. Find the common difference of the following arithmetic sequences.

a) 2, 4, 6, 8,

b) 5, 3, 1, -1,

c) , 3/2 , 5/2 ,

Solutions:

a) d = T2 T1 = 4 2 = 2

d = T3 T2 = 6 4 = 2

b) d = T2 T1 = 3 5 = -2

d = T3 T2 = 1 3 = -2

c) d = T2 T1 = 3/2 = 1

d = T3 T2 = 5/2 3/2 = 1

The n-th term can be calculate by using the given formula. Tn = a + (n 1)d

Example:

1. Find the tenth term of the following sequence. 5, 10, 15, 20, 25,.

solution:

a = 5 & d = 10 5 = 5

Tn = a + (n 1)d

T10 = 5 + (10 1)5

= 5 + (9)5

= 5 + 45

= 50

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2. Find the:

a) common difference, d

b) the 11th term of the arithmetic sequence

3, 6, 9, 12, 15, 18

Solution:

a) d = T2-T1 = 6 3 = 3

b) T11 = a + (n-1)d

= 3 + (11-1)(3)

= 33

3. Find the number of terms that exist in the following arithmetic progression.

3, 5, 7,.,81

Solution:

a = 3, d = 2 , n = ? & Tn=81

Tn = 81

a + (n 1)d = 81

3 + (n 1)2 = 81

3 + 2n 2 = 81

2n = 81 3 + 2

n = 80 2

n = 40

4. Given that term 7 of an arithmetic sequence is 49 and term 16 is 112. Find the sequence.

... 42, 35, 28, 21, 14, 7, is sequence The

7

4942

49)7(6

49)(6

:(1) into 7

7

9

63

639:)1()2(

211215

112)116(

1496

49)17(

112

49

16

16

7

7

16

7

a

a

a

da

dSubstitute

d

d

d

daT

daT

daT

daT

T

T

1. Find the common difference and the tenth term of the following.

i) 3, 7, 11, 15,

ii) 1, 2.5, 4, 5.5, .

iii) 9, 6, 3, 0, .

2. Find the number of terms that exist in the following arithmetic progression.

i) 2, 5, 8, , 59.

ii) 7, 15, 23, , 103.

iii) 6, 19, 32, , 318.

3. The 10th term of the arithmetic sequence

a, a+3, a+6, is 34. Find the value of a and the 20th term.

4. Given that the first term of an arithmetic progression is 6 and the 8th term is -29. Find the sixteenth term.

5. Given the value of the third term of an arithmetic sequence is 5 and the value of the seventh term is -1. Find the common difference and the first term.

To calculate the sum of the first n-th terms, we need to determine the first term (a), common difference (d) and the number of term for the given sequence.

The formula of the sum of the first n-th term:

dnanSn )1(2

2

3

Example: 1. Find the sum of the first fifth teen terms for the given

sequence. 21, 19, 17, 15,5 solution:

a = 21 , d = -2 & S15 = ? S15 = 15/2 [2(21) + (15 - 1)(-2)] = 15/2 [ 42 + 14(-2) ] = 15/2 [ 42 -28 ] = 15/2 [14 ] = 105

dnanSn )1(22

2. Determine the sum of the sequence

57, 63, 69, 75, 81,.., 177.

Solution:

a = 57 , d = 6 , n = ? & Sn = ?

first we need to find the number of the sequence.

Tn = 177

a + (n 1)d = 177

57 + (n 1)6 = 177

57 + 6n 6 = 177

6n = 177 57 + 6

n = 126 6

n = 21

Now, we know that n = 21, so we could determine the S21 using the sum formula.

S21 = 21/2 [ 2(57) + (21 1) 6 ]

= 21/2 [ 114 + 120 ]

= 21/2 [ 234 ]

= 2457

dnanSn )1(22

1. Find the sum of the first 13 terms of the arithmetic progression: 5, 10, 15, .

2. Find the sum of the sequence; 101, 103, 105, , 115.

3. Find the sum of the sequence; 5, 3.5, 2, , -10.

4. The first term of an arithmetic progression is 5 and the 20th term is 62. Find the sum of the first 20 terms of this progression.

5. The 4th term of an arithmetic progression is 8. Given that the sum of the first 12 terms is 126. Find the sum of the next 8 terms.

6. The first term of an arithmetic progression is -27. If the 10th term is equal to the sum of the first 9 terms of this progression, find the common difference and the sum of the first 20 terms.

Ishak starts with a monthly salary of RM 1250 for the first year and receives an annual increment of RM 80. How much will he receive monthly for his tenth year of service?

solution:

a = 1250 ,d = 80 , T10 = ?

Tn = a + (n 1)d

T10 = 1250 + (10 1)80

= 1250 + (9)80

= 1250 + 720

= RM 1970

Geometric Sequence

Definition

Sequence of numbers whereby each term(except the first term) is obtained by multiplying a constant

called the common ratio(r) to its immediate predecessor.

Common Ratio

The common ratio can be either positive or negative value.

Common Ratio, r = T2 T1

The nth term

Tn = arn-1

Sum of the first n terms 1;

1

)1(

r

r

raS

n

n1;

1

)1(

r

r

raS

n

n

4

To detemine whether given sequence is a geometric sequence or not, we need to find for more than one common ratio of the sequence.

If all the common ratio are same then it is a geometric sequence.

Example:

1. State whether the sequences below are geometric sequence or not

a) 15, 45, 135, 405, 1215,

b) 4, 16, 64, 256, 1024,

Solution:

a) r = T2 T1 = 45 15 = 3

r = T3 T4 = 135 45 = 3 So, it is a geometric sequence.

b) r = T2 T1 = 16 4 = 4

r = T3 T4 = 64 4 = 4 So, it is a geometric sequence.

The n-th term can be calculate by using the given formula.

Tn = arn-1

Example:

1. Find the 10th term of the sequence: -1, 2, -4, 8, .

solution:

a = -1 , r = -2 &

Tn = arn-1

T10 = (-1)(-2)10-1

= (-1)(-2)9

= (-1)(-512)

= 512

2. Find the number of terms that exists for the geometric progression: 8, 4, , 1/32.

Solution:

a = 8 , r = , Tn = 1/32 & n = ?

Tn = arn-1

1/32 = (8)(1/2)n-1

1/32 8 = (1/2)n-1

1/32 8 = (1/2)n-1

1/256 = (1/2)n-1

(1/2)8 = (1/2)n-1

So, 8 = n - 1

n = 9

9

18

)1(

)2

1log(

256

1log

)2

1log()1(

256

1log

)2

1log(

256

1log 1

n

n

n

n

n

Using log

3. Given that the second term of a geometric sequence is -18 and the fourth term is -2. Find the common ratio and the first term.

Solution:

18

182

ar

T

2

23

4

ar

T1 2

3

1

9

1

9

1

18

2

2

3

r

r

ar

ar2 1

If r = -1/3, then ar = -18

a (-1/3) = -18

a = -18 (-1/3)

a = 54

And

If r = 1/3, then ar = -18

a (1/3) = -18

a = -18 (1/3)

a = -54

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Solve the following:

1. State whether the sequence below are geometric sequence or not.

i)

ii) 0.1, 0.5, 1, 5,

2. Find the 15th term of the sequence 2, 4, 8, 16, 32, .

3. Given a geometric sequence of 1/9, 1/3, 1, , 81. Find:

i) the 5th term of the sequence

ii) number of terms that exist in the sequence.

,...3,1,3

1,

9

1,

27

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Sum of the First n-th Terms

To calculate the sum of the first n-th terms, we need to determine the first term (a), common difference (d) and the number of term for the given sequence.

The formula are depend on the value of the common ratio.

1; 1

)1(

r

r

raS

n

n1;

1

)1(

r

r

raS

n

n

1. Find the sum of the first 8 terms of the geometric sequence: 8, -24, 72, .

Solution:

a = 8 , r = -3 (is < 1) & S8 = ?

S8 =

=

1; 1

)1(

r

r

raS

n

n

13120

)3(1

))3(1(8 8

Solve the following: 2. Find the sum of the first 10 terms of the geometric

sequence 4, 8, 16, .

3. Find the sum of the geometric progression below. 10, 5, 2.5, , 5/32.

4. The sum of the first 5 terms of a geometric sequence is 211/27 and the common ratio is 2/3. find the first term of this geometric sequence.

5. Given the sum of the 1st term and 3rd term of a geometric sequence is 9/2. The sum of the 2nd and the 4th term is 27/4. find the common ratio and the 1st term.

1. Aida saves RM 1000 in a saving account that pays 8% compounded annually. Find the amount in her account at the end of 5 years.

solution:

a = 1000 , r = 100% + 8% = 108% = 1.08 & T5 = ?

Tn = arn-1

T5 = (1000)(1.08)4

= RM 1360.49

The amount in the account at the end of 5 years is RM1360.49

2. Mrs. Gan estimates that the monthly expenses of a baby will increase by 4% every month. If the expenses for the first month are estimated to RM 50, determine

a. type of sequences that can be used to estimate the monthly expenses

b. The expenses for the 25th month

c. the total expenses for the first 12 months

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Tutorial 1

Quiz 1