progressions

There are three types of progressions Arithmetic progression Geometric progression Harmonic progression

Introduction

A sequence is said to be in Arithmetic Progression when they increase or decrease by a constant number. This constant number is called the common difference (c.d) of the arithmetic progression.

Example: 1, 3, 5, 7, ………...... c.d. = 2 –7, –3, 1, 5, 9 ……… c.d. = 4 8, 5, 2, –1, –4 ……… c.d. = –3

Arithmetic progression

Tn = a + (n-1)d

First term

Number of terms

Common difference

Nth term of AP

Sn= n/2[2a + (n-1)d]

Sum to n terms

Number of terms

First term

Common difference

Given that first term of a series is 7 and the last term of the same to be 49. The sum of the series is 336. Find the number of elements in this series and the common difference.

a = 7 Tn =49 Sn = 336 n=? d=?

Problem 1

Tn = a + (n-1)dÞ 49 = 7 + (n-1)d ……………………………(1)Þ Sn = n/2[2a + (n-1)d]Þ 336 = n/2[2*7 + (n-1)d]Þ 336 = n/2[14 + 42] ( from equation 1)Þ 336 = 28nÞ n= 336/28 = 12Þ Substituting the value of n in equation 1Þ 49 = 7 + 11dÞ d = 42/11

A gentleman buys every year Bank's cash certificates of value exceeding the last year's purchase by Rs. 300. After 20 years, he finds that the total value of the certificates purchased by him is Rs. 83,000. Find the value of the certificates purchased by him in the 13th year.

Problem 2

Given: d = 300 n = 20 Sn = 83000 Sn = n/2[2a + (n-1)d]Þ 83000 = 10[2a + 19*300]Þ 83000 = 10[2a + 5700]Þ 8300-5700 = 2aÞ 2600/2 = aÞ a = 1300

the value of the certificates purchased by him in the 13th year = 13th term

13th term = a + (13-1)d=> 13th term = 1300 + 12*300Þ 13th term = 4900Þ the value of the certificates purchased by

him in the 13th year = 4900

The sum of the first and the 9th term of an arithmetic progression is 24. What is the sum of the first nine terms of the progression?

Given: 1st term + 9th term = 24Þ a + a + 8d = 24

Þ 2a + 8d = 24…………..1Þ S9 = 9/2[2a + 8d]Þ S9 = 9/2[24]Þ S9 = 108

Problem 3

The sum of the first 50 terms common to the Arithmetic Sequence 15, 19, 23..... and the Arithmetic Sequence 14, 19, 24..... is.

Given: series 1 : 15, 19, 23…….. a = 15, d = 4 series 2 : 14, 19, 24……. a = 14, d = 5

Problem 4

The series whose terms are common to both will also be an arithmetic series with common difference = LCM of the common differences of the two given series

so the common difference of the required series = 20

First term = 19 Sum to 50 terms = 25[2*19+49*20]=> Sum to 50 terms = 25[38+980]=> Sum to 50 terms = 25[38+980] =

25450

How many terms should be taken of the series 42, 39, 36….. To make 315.

Given: series 42, 39, 36 …….. Since 39 – 42 = 36 – 39 = -3 so the series is

arithmetic. First term a = 42 Common difference d = -3 Sum to n terms S = 315 n = ?

Problem 5

S = n/2[2a + (n-1)d]Þ 315 = n/2[2*42 + (n-1)*(-3)]Þ 315 = n/2[84 -3n + 3]Þ 315 = n/2[87 -3n]Þ 630 = n [87 -3n]Þ 630 = [87n - 3n2]Þ 3n2 – 87n + 630 = 0Þ n2 – 29n + 210 = 0Þ n2 – 14n – 15n + 210 = 0Þ n(n-14) -15(n- 14) = 0Þ (n-14) (n-15) = 0, n= 14 or 15

The sum of 5 numbers in AP is 75 and the product of the greatest and the least is 161. Find out the greatest number of the series.

Let the five numbers be a-2d, a-d, a, a+d and a+2d

Sum = 75=> a-2d + a-d + a + a+ d +a+2d = 75Þ 5a = 75Þ a = 15

Problem 6

Product of the greatest and the least = 161

Þ (a + 2d) (a – 2d) = 161Þ (15 + 2d) (15 – 2d) = 161Þ 225 - 4d2 = 161Þ 4d2 = 225-161Þ 4d2 = 64Þ d2 = 16Þ d = 4Þ greatest number = a + 2d = 15 + 2*4 =

23

The first term and the fifth term of an arithmetic progression containing of five terms are 2 and 14 respectively. The first term and fourth term of another arithmetic progression containing four terms are 3 and 9 respectively. If all terms in both the series are added together then their sum will be equal to

Problem 7

Series 1 a = 2 5th term = a + 4d =

14 d= 3, n = 5 Sum to 5 terms =

5/2[2a + 4d]Þ Sum to 5 terms =

5/2[4 + 12]Þ Sum to 5 terms = 40

Series 2 a = 3 4th term = a + 3d

= 9 d = 2, n = 4 Sum to 4 terms =

4/2[2a + 3d]=> Sum to 4 terms

= 2[6+6]= 24

Required sum = 40 +24 = 64

ABC transformers has produced 780 transformers in 2008 and is decreasing the annual production by 40 transformers per year because of the competition by Class transformers. Class transformers has produced 100 transformers in 2008 and is increasing the annual production by 30 transformers. In which year Class transformers will become the larger producer?

Problem 8

Let by nth year Class transformers will become larger producer

Then nth term of Class transformer > nth term of ABC transformers

For Class transformer a = 100, d = 30 For ABC transformer a = 780, d= -40 Þ 100 + (n-1) 30 > 780 + (n-1) (-40)Þ 100 + 30n -30 > 780 – 40n + 40Þ 70n > 750Þ n > 750/70Þ n> 10.71Þ From 11th year (2019) Class transformer

will be the larger producer.

A machine is bought for Rs. 90000 and the depreciation on it is assessed at Rs. 7200 per year. What will be its book value at the end of eight years?

Given: a =90000 Depreciation d = -7200 n = 9 9th term = a + 8d 9th term = 90000 – 8*7200 9th term = Rs. 32400

Problem 9

Find the annual salary increment of a man who retires after his 37th year , having earned Rs. 61700 in his fifth year and an average of Rs. 100900 over his whole career.

Given: 5th term = 61700Þ a + 4d = 61700………………….1Þ Average salary over his whole career =

100900

Problem 10

Þ Total salary earned = average salary * years of employment

Þ total salary earned = 100900*37 = 3733300

Þ 37/2[2a + (37-1)d] = 3733300Þ 2a + 36d = 3733300*2/37Þ a + 18 d = 100900……………..2Þ By solving equations 1 and 2Þ d = 2800Þ Annual salary increment of the person =

Rs. 2800

The salary of a company secretary is increased by a fixed increment each year. If his total earnings over nine years are Rs. 2340000 and his salary in the final year is Rs.295000, what was his salary in the sixth year?

Since the salary is increased by a fixed increment each year so the series is AP

Sum to 9th term = 2340000 9th term = 295000

Problem 11

Þ 9/2[2a +8d] = 2340000Þ a + 4d = 260000……………1Þ a+ 8d = 295000……………..2 Þ By solving equation 1 and 2Þ d = 8750, a = 225000Þ Salary in the 6th year = a + 5dÞ Salary in the 6th year = 225000 + 5*8750Þ Salary in the 6th year = Rs. 268750

Geometric progression (also inaccurately known as a geometric series) is a sequence of numbers such that the quotient of any two successive members of the sequence is a constant called the common ratio of the sequence.

A sequence with a common ratio of 2 and a scale factor (first term) of 1 is1, 2, 4, 8, 16, 32...

A sequence with a common ratio of -1 and a scale factor (first term) of 5 is5, -5, 5, -5, 5, -5,...

Geometric progression

If the common ratio is: Negative, the results will alternate between positive

and negative. Greater than 1, there will be exponential growth

towards infinity (positive). Less than -1, there will be exponential growth towards

infinity (positive and negative). Between 1 and -1, there will be exponential decay

towards zero. Zero, the results will remain at zero

Tn = arn-1

Nth termFirst term

Common ratio Number of terms

If common ratio > 1 Sum to n terms sn = a(rn -1)/(r-1)

If common ratio < 1 Sum to n terms sn = a(1 -rn)/(1- r)

Sum to infinity when r<1 S = a/(1-r)

Summation

Find the cost of an annuity of Rs. 25000 for 15 years, if compound interest is allowed at 3% per annum.

Given: a =25000 n = 15 years r = 3% =1.03 P = a(rn – 1)/rn(r-1) P = 25000(1.0315 – 1) /1.0315(1.03-1)

Problem 12

A firm borrows Rs. 1000000 and repays it by three equal sums at the ends of the following three years. What is the amount of each repayment, if compound interest at 41/2% is allowed?

Given: p =1000000 n =3 r= 1.045 a = ? P = a(rn – 1)/rn(r-1) a=p rn(r-1) (rn – 1)

Problem 13

What sum should be paid for an annuity of Rs. 25000 per annum, to be paid half-yearly commencing in six months’ time, if compound interest is allowed at 14% per annum and compounded half yearly and the annuity is to last

(A) 10 years (B) 20 years (C) What is the amount payable in the

above case in perpetual case?

Problem 14

Given: a= Rs. 25000 Compound interest 14% per annum compounded half

yearly (14/2) r = 1.07 (A) Time = 10 years n = 20 P = a(rn – 1)/rn(r-1) P = 25000(1.0720 – 1)/1.0720(1.07-1)

(B) Time = 20 years n = 40 P = a(rn – 1)/rn(r-1) P = 25000(1.0740 – 1)/1.0740(1.07-1) (C) for Perpetual Annuity sum to infinity is used P = a /(r-1) P = 25000/(1.07-1)

A small water pump costs Rs. 6200 and is expected to last for 14 years and then have a scrap value of Rs. 740. If depreciation is to be calculated as a fixed percentage of the current book value at the end of each year, what should the percentage be?

Since depreciation is to be calculated as a fixed percentage of the current book value so the series is GP

Problem 15

Given: a = Rs. 6200 n = 15 15th term = 740 r = ? ar14 = 740 r14 = 740/6200 = 0.11935

Matsushita Electronics Ltd is a reputed manufacturer of TV sets sold as a brand name of ‘Panasonic’. It has produced 150000 TV sets in India five years back. If the total production of the company today, at the end of 5th year of operations in India, is 800000 TV sets, then:

A. Estimate by how many units, production has increased each year if the increase in terms of number of units is the same each year.

Problem 16

B. Estimate by how many units production has increased each year if the increase in terms of percentage of last year’s production is the same every year.

C. Based on the annual increment in production in both fixed units and percentage terms, forecast the amount of production for the 10th year.

D. If the total market size is 3000000 TV units today and is growing by 5 percent per annum, in how many years would Matsuhita grab 50 percent of the market if it keeps growing at the same percentage rate calculated earlier?

Given: Production of TV before 5 years = 150000

Þ a = 150000 Þ 5th term = 800000

A. increase in terms of number of units is the same each year.

Þ The series is APÞ 5th term = 800000Þ a + 4d = 800000Þ 150000 + 4d = 800000Þ 4d = 650000 => d = 162500

B. The increase in terms of percentage of last year’s production is the same every year.

The series is GPÞ a = 150000 Þ 5th term = 800000Þ ar4 = 800000Þ r4 = 800000/150000Þ r =1.52(approx)

C . For fixed units of increment (AP)10th year’s production = a + 9dÞ 10th year’s production = 150000 + 9*

162500Þ 10th year’s production = 1612500 For percentage increment(GP)10th year’s production = ar9

10th year’s production = 150000*1.529

10th year’s production = 6496561.44

D. Total market size =a =3000000 growing by 5% per annumÞ r = 1.05Þ Let after n years Matsuhita will grab 50

percent of the marketÞ Then nth term of Matsuhita = 50% of nth

term of the market sizeÞ 800000*1.52n = ½ * 3000000*1.05n

Þ 1.52n /1.05n= ½ * 3000000/800000Þ 1.447n = ½ * 3000000/800000Þ 1.447n = 1.875

Find out the sum of an infinite geometrical series whose first three terms are 10, 5 and 2.5 respectively.

Given: Infinite geometric series 10, 5, 2.5 ……

a = 10, r = 5/ 10 = ½ Sum to infinity = a/ (1-r) Sum to infinity = 10/(1-1/2) Sum to infinity = 20

Problem 17

Find the geometric series in which the 10th term is 320 and 6th term is 20.

10th term = 320Þ ar9 = 320…………………1Þ 6th term = 20Þ ar5 = 20……………………..2Þ Dividing equation 1 by equation 2Þ ar9

/ar5 = 320/20

problem18

Þ r4 = 16Þ r = 2Þ Substituting the value of r in equation 2Þ a25 = 20Þ a*32 = 20Þ a = 20/32 = 5/8Þ So the series is 5/8, 5/4, 5/2, 5 …..

Company has bought a printer for Rs.10000. This machine is expected to be utilized for the next 5 years and the salvage value at the end of five years is expected to be RS. 1000 only.

Calculate the depreciation by straight method

Calculate the depreciation by Reducing balance method

Problem 19

Annual depreciation = (total cost – salvage value)/ no. of years

Annual depreciation = (10000 -1000)/5 = Rs. 1800 per year Rate of depreciation = x X% of depreciable cost = 1800 X% of (10000-1000) = 1800 X = 20%

Straight line method

Period Depreciable cost

Depreciation rate

Depreciation expense

Accumulated depreciation

Book value

2011 9000 20% 1800 1800 8200

2012 9000 20% 1800 3600 6400

2013 9000 20% 1800 5400 4600

2014 9000 20% 1800 7200 2800

2015 9000 20% 1800 9000 1000

Written down value method (Reducing Balance Method) This method uses depreciation rate which is

twice the straight line depreciation rate.

Straight line depreciation rate = 20%

So reducing balance depreciation rate = 40 %

Period Depreciable cost

Depreciation rate

Depreciation expense

Accumulated depreciation

Book value

2011 10000 40% 4000 4000 6000

2012 6000 40% 2400 6400 3600

2013 3600 40% 1440 7840 2160

2014 2160 40% 864 8704 1296

2015 1296 22.83% 296 9000 1000

Problem 20 A stamping machine bought for Rs. 50000 is

expected to be utilized for the next 10 years. The salvage value at the end of 10 years is expected to be Rs. 20000 Calculate the depreciation

(A) By straight line depreciation method (B) By reducing balance method

Annual depreciation = (total cost – salvage value)/ no. of years

Annual depreciation = (50000 -20000)/10 = Rs. 3000 per year Rate of depreciation = x x% of depreciable cost = 3000 x% of (50000-20000) = 3000 x = 10%

Period Depreciable cost

Depreciation rate

Depreciation expense

Accumulated depreciation

Book value

2011 30000 10% 3000 3000 47000

2012 30000 10% 3000 6000 44000

2013 30000 10% 3000 9000 41000

2014 30000 10% 3000 12000 38000

2015 30000 10% 3000 15000 35000

2016 30000 10% 3000 18000 32000

2017 30000 10% 3000 21000 29000

2018 30000 10% 3000 24000 26000

2019 30000 10% 3000 27000 23000

2020 30000 10% 3000 30000 20000

Written down value method (Reducing Balance Method) This method uses depreciation rate which is

twice the straight line depreciation rate.

Straight line depreciation rate = 10%

So reducing balance depreciation rate = 20 %

Period

Depreciable cost

Depreciation rate

Depreciation expense

Accumulated depreciation

Book value

2011 50000 20% 10000 10000 40000

2012 40000 20% 8000 18000 32000

2013 32000 20% 6400 24400 25600

2014 25600 20% 5120 29520 20480

2015 20480 2.34% 480 30000 20000

2016 20000 -- -- 30000 20000

2017 20000 -- -- 30000 20000

2018 20000 -- -- 30000 20000

2019 20000 -- -- 30000 20000

2020 20000 -- -- 30000 20000