SEQUENCE AND SERIES

SEQUENCE: - A Sequence is an ordered set of objects formed by a particular rule and which are functions of a particular numbers . It can otherwise be set of numbers generated in accordance with a definite pattern constitute sequence.

Consider each of the following sets of number

(a) 5, 9, 13, 17, 21,- , - ,-

(b) 2, 4, 6, 8, 10,-, -, -,

Each member of each set is called a term .The dots after each set shows that set of numbers continues indefinitely.

Can you identify the pattern each set of number above and general formula for the general term of the sequence?

Example: Find the first three terms of the sequence whose general term is

1. T_n= 3 + (-1)ⁿ

2. T_n =

Solution

(a) T_n= 3 + (-1)ⁿ

T₁ = 3 + (-1)¹ =2

T₂ = 3 + (-1)² =4

T₃ = 3 + (-1)³ =2

(b)

THE N^TH TERM OF AN ARITHMETIC PROGRESSION

Consider a linear sequence, with a first term a, and common difference d, we wish to find n^th term T_nof the linear sequence.

T₁ = a

T₂ = a + d

T₃ = a + 2d

T₄ = a + 3d

T_n = a + (n– 1)d

Hence n^thterm of an Arithmetic progression T_n whose first term is a and common difference d, is given by the formula

T_n = a + (n – 1) d.

Example

Find the 20^th term of the linear sequence

6, 12, 18, 24, – – –

Solution

a = 6, d = 12 – 6 = 6, n = 20

T_n = a + (n – 1)d

T₂₀ = 6 + (20 – 1)6

T_{20 =}6 + 114

T₂₀ = 120

Example

The first term of a linear sequence is 3 and the 8^th term is 31. Find the common difference

Solution

a = 3, n = 8, T₈ = 31

T_n = a + (n – 1)d

T₈ = 3 + (8 – 1)d = 31

3 + 7d = 31

7d = 28

d = 4

Example

The 8^th term of a linear sequence is 18 and the 12^th term 26. Find the first term, the common difference and the 20^th term.

Solution

T₈ = 18 = a + 7d – – – – – – – – – (i)

T₁₂ = 26 = a + 11d – – – – – – – –(ii)

Subtract (2) from (i)

–8 = –4d

d = 2

Substituting d = 2 into (i)

18 = a + 7 (2)

a = 4

T₂₀ = 4 + (20 – 1) 2

T₂₀ = 4 + 38=42

Example

Find the value of x, given that x+2, 3x, 4x + 2 are consecutive term of a arithmetic sequence

Solution

If x + 2, 3x, 4x + 2 form an arithmetic sequence,

d = 3x – (x + 2)

= 2x – 2 – – – – – – – – – – (i)

Also

d = 4x + 6 – 3x

= x + 6 – – – – – – – – –(ii)

Equating (i) and (ii) {since the common difference must be equal}

2x – 2 = x + 6

x = 8

ARITHMETIC MEAN

The Arithmetic mean of two numbers a and c is the number b between a and c in the arithmetic progression.

Thus, if a, b, c, form an arithmetic sequence

Then, common difference

d = b – a – – – – – – – – – – – – – – – (i)

Also d = c – b – – – – – – – – – – – – – – – (ii)

Equating (i) and (ii)

b – a = c – b

Example

Find the arithmetic mean of 4 and 16

Solution

Let the arithmetic mean be b

Then 4, b, 20

Example

Insert four arithmetic mean between 5 and 80

Solution

Let the means be denoted by A, B, C, D then 5, A, B, C, D, 80 are in Arithmetic Progression

a = 5, n = 6, T₆ = 80

T₆ = 80 = 5 + (6 – 1)d

80 = 5+5d

75 = 5d

d = 15

T₂ = a + d

= 5 + 15 = 20

T₃ = 5 + 2 × 15 = 35

T₄ = 5 +3 × 15 = 50

T₅ = 5 + 4 × 15 = 65

Example:– If x², y², z² are in AP, show that

Example

The first and the last term of an A.P. are 20 and 110 with a common difference. How many terms has the A.P

Solution

a = 20

l = 110

d= 15

T_n = l= a + (n – 1)d

110 = 20 + (n – 1) 15

110 = 20 + 15n – 15

105 = 15n

n = 7

Thus, the A.P. has 7 terms

SERIES

A series is formed by the sum of the terms of a sequence.

e.g 1, 5, 9, 13, 17, – – – – is a sequence

but 1 + 5 + 9 + 13 + 17 + – – – – is a series. The Sum: 1 + 5 + 9 + 13 + 17 + – – – is designated by S_n

i.e. S_n= 1 + 5 + 9 + 13 + 17 + – – –

Consider the sequence, T₁ + T₂ + T₃ + T₄+ – – – – – – is the corresponding series.

If the sequence contains finite number of terms, the corresponding series is infinite.

The sum

S_n = T₁, + T₂, + T₃, + T₄, + – – – – – – – – + T_n

S_n+1 = T₁, + T₂, + T₃, + T₅, + – – – – – – – –– T_n + T_n+1

Example

The sum of the 4^th and 6^th term of arithmetic is 42. The sum of the 3^rd and 9^th term of the progression is 52. Find the first term, the common difference and sum of the first four term.

Solution

T₄ = a + 3d

T₆ = a + 5d

T₄ + T₆ = a + 3d + a + 5d = 42

= 2a + 8d = 42 – – – – – – – (i)

T₃ = a + 2d

T₉ = a + 8d

T₃ + T₉ = a + 2d + a + 8d = 52

= 2a + 10d = 52 – – – – – – (ii)

Subtract (i) from (ii)

2d = 10

d = 5

Substituting d = 5 into (i)

2a + 8 (5) = 42

2a + 40 = 42

a = 1

Therefore the terms will be the following

T₁ = 1

T₂ = a + d = 1 + 5 = 6

T₃ = a + 2d = 1 + 2(5) = 11

T₄ = a + 3d = 1 + 3 (5) = 16

S₄ = T₁ + T₂ + T₃ + T₄

S₄ = 1 + 6 + 11 + 16

S₄ = 34

SUM OF N TERMS OF NTH ARITHMETIC PROGRESSION

The general arithmetic series can be written as

S_n = a +(a + d) + (a + 2d) + (a + 3d) + – – –+( a + (n – 1)d)

Rewriting the term in reverse order

S_n = (a + (n – 1) d )+( a + (n – 2) d) + (a + (n – 3) d) + – – – + a

Adding the two lines together

2S_n = 2a + (n – 1) d + 2a + (n – 1) d + – – – – – – – 2a + (n – 1)d

There are n terms in this series

2S_n = n (2a + (n – 1)d

S_n = (2a + (n – 1)d

Sum of an A.P. = [2a + (n – 1)d]

Or (a +l)

Where l = t_n = a + (n – 1)d

Example

Find the sum of the first twelve terms of the sequence 2, 5, 8, 11, – – –

Solution

a = 2

d = 5 – 2 = 3

n = 12

S_n = [2a +(n – 1)d]

S₁₂ = [2 × 2 + (12 – 1) 3]

S₁₂ = 6 (4 + 33)

S₁₂ = 222

Example

The sum of the first ten terms of an A.P. is – 60 and the sum of the first fifteen terms the sequence is – 165. Find the 18^th term the sequence.

Solution

S_n = [2a + (n – 1)d]

S₁₀ = [2a + 9d]

5(2a + 9d) = – 60

2a + 9d = –12 – – – – – – – – – – – – – – (i)

S₁₅ = [2a + 14d] = – 165

2a + 14d = –22 – – – – – – – – – – – – – –(ii)

Subtract (i) from (ii)

5d = –10

d = –2

Substitute d = –2 into equation (i)

2a + a (–2) = –12

2a = –12 + 18

a = 3

The 18^th term will be

T₁₈ = 3 + (18 –1) –2

T₁₈ = 3 – 34

T₁₈ = –31

Example

The sum of an A.P. is 950, the first term being 2 and common difference 3, find the number of terms in the series.

Solution

S_n= [2a +(n–1)d]

S_n = 950, a = 2, d = 3

950 = [2 ×2 + (n – 1)3]

950 = [4 + 3n – 3]

950 = (1 + 3n)

1900 = n + 3n²

3n² + 76n – 75n – 9500 = 0

n(3n + 76) – 25 (3n + 76) = 0

(n – 25) (3n + 76) = 0

n = 25 n {n cannot be negative integer}

n = 25

Example

The sum of three numbers in A.P is 15 and their produce it 80. Find the numbers.

Solution

Let the terms be a – d, a, a + d

Sum of the terms = (a – d) + a + (a + d) = 15 – – – – – – – – – – – – – (1)

3a = 15

a = 5

Also, Product: (a – d) × a × (a + d) = 80

(a² – d²) a = 80 – – – – – – – – – – – – – – – – – (2)

Substituting a = 5 into (2)

(5² – d²) 5 = 80

25 – d² = 16

25 – 16 = d²

9 = d²

d = + 3

Thus the series (5 – 3) + 5 + (5 + 3) = 2 + 5 + 8

Example

The sum of four integers in A.P. is 20 and the sum of their sequence is 120. Find the numbers (Hint, select the number to be a – 3d, a – d, a + d, a + 3d)

Solution

Let the terms be a – 3d, a – d, a + d, a+3d

Sum of the terms = (a – 3d) + (a – d) + (a + d) + (a + 3d) = 20

4a = 20

a = 5

Sum of the squares of the term

= (a – 3d)² + (a – d)² + (a + d)² + (a + 3d)² = 20

a² – 6ad + 9d² + a² – 2ad + d² + a² + 2ad + d² + a² + 6ad + 9d² = 120

4a² + 20d² = 120 – – – – – – – – – – – – (a)

Substituting a = 5 into (a)

4(5²) + 20d² = 120

100 + 20d² = 120

d = + 1

The series = (a – 3d) + (a – d) + (a + d) + (a + 3d)

= (5 – 3) + (5 – 1) + ( + 1) + ( + 3)

= 2 + 4 + 6 + 8

Example

The sum of five number in A.P. is 20 and their product is 720. Find the number.

Solution

Let the terms be (a – 2d), (a – d), a, (a +d), (a + 2d)

Sum of the numbers = a – 2d + a – d + a + a + d + a + 2d = 20

5a = 20

a = 4

Product of the numbers = (a – 2d) (a – d) (a + 2d) a = 720

= (a² – 4d²)(a²–d²) a = 720

Substituting a = 4 in the product of the numbers

(4² – 4d²)(4² – d²)4 = 720

(16 – 4d²)(16 – d²) = 180

256 – 16d² – 64d² + 4d⁴ = 180

256 – 80d² + 4d⁴ = 180

64 – 20d² + d⁴ = 45

19 – 20d² + d⁴ = 0

Let d² = p

19 – 20p + p² = 0

(p – 19) (p – 1) = 0

p = 1, p = 19

d² = 1 or d² = 19

d = + 1 or d = +

d = + 1

Therefore d = 1

The series is 2, 3, 5, 6,

Note:– It is advisable when dealing problem involving three (five or any odd number of terms in A.P. to make sure the middle term is a

Example: Show the sum of n term of the progression logx, logx², logx³, logx⁴ – – – – – is logx

Solution S_n = logx + logx² + logx³ + logx⁴ + – – – – – – + logxⁿ

S_n = logx + 2logx + 3logx + 4logx + – – – – – nlogx

a = logx

d = 2 logx – logx = logx

S_n = [2a +(n – 1)d]

= [(2log + (n –1)logx]

= [(2 + n – 1) logx]

= (n+1) logx

Example

Find the sum of all numbers between 200 and 500 which are divisible by 17

Solution

Since 13 is left as remainder on division of 200 by 17, the least number greater than 200 divisible by 17 is 204. Again if we divide 500 by 17, 7 is left as remainder. This implies that greater number less than 500 which is divisible by 17 is 493 .

Thse are 204, 221, 238 – – – – 493

a = 204, d = 17, l = 493

l = a + (n – 1) d

493 = 204 + (n – 1) 17

493 = 204 + 17n – 17

493 = 187 + 17n

There are 18 numbers between 200 and 500 which are divisible by 17

APPLICATION OF ARITHMETIC PROGRESSION TO BUSINESS

Example

The Chief Accountant of O.A.U. Community Bank Ltd Pay out commission in the form of A.P. N200, N500, N800 – – – – –– – to people

If he finds out that he paid out N15500. How many people has he paid?

Solution

a = N200 , d = N300 Sn = N15500

S_n = [2a +(n – 1)d]

15500 = [2 x 200 + (n – 1) 300]

15500 = (400 + 300n – 300)

15500 = (100 + 300n)

31000 = 100n + 300n²

310 = n + 3n²

3n² + n – 310 = 0

3n² + 31n – 30n – 310 = 0

n(3n + 31) – 10(3n + 31) = 0

(n – 10) (3n + 31) = 0

n = 10 or n =

The Chief Accountant must have paid 10 people since n cannot be a negative value.

Example

The savings of a trader is given as N125, N150, N175, N200, – – – – calculate

(a) The trader twentieth saving

(b) The total money the trader would collect at the twentieth saving.

Solution

a = N125, d = N25 n = 20

T_n = a + (n – 1)d

T₂₀= 125 + (20 – 1) 25

T₂₀ = 125+ 475

T₂₀ = 600

The twentieth savings is N600

(b) The total money for the twentieth saving

S_n = (a + 1)

S₂₀ = (125 + 600) = 10 (725)

S₂₀ = 7250

Example

Two posts were offered to a fresh graduate of Demography. In one, the starting salary was N24,000 per month and the annual increment of N1,600. In the other post, the salary commences at N240,000 per annum. The annual increment was N2,400. The demographer decided to accept the post which would give him more earning in the first 25 years of services. Which was acceptable to him? Justify your answer.

Solution

For the first job

Annual Salary = N24,000 × 12 = N288,000

Total sum that he would collect after 25 years will be

S₂₅ = [2 × 288,000 + (25 – 1) 1600]

S₂₅ = 12.5 [576,000 + 38400]

S₂₅ = 12.5 × 614,400

S₂₅ = N7,680,000

For the second job

1st annual salary = a = N204,00

Annual increment = d = N2,400

Total sum collected after 25 years will be

S₂₅ = [2 × 204,000 + (25 –1) 2,400]

S₂₅ = 12.5 [408,000 + 57600]

S₂₅ = 12.5 × 465,600

S₂₅ = N5,820,000

From our calculation, it shows that the Demographer accepted the first job

Example

The rate of monthly salary of Mr Jejelayegba increased annual in A.P. It is known that he was drawing N400 a month during the 11th year of his service and N760 during 29th year rate of annual increment. What should be his salary at the time of retirement just on the 36 years of services?

Solution

Monthly salary of the man in the 11th year = N400

Annual salary will be N400 x 12 = N4800

Monthly salary of the man in the 29th year = N760

Annual salary will be N760 x 12 = N9120

Since T_n = a + (n – 1)d

In the 11th year

T₁₁ = a + (11 – 1)d = N4800

a + 10d = 4800 – – – – – – – – – – – – – (1)

In the 29th year

T₂₉ = a + (29 – 1)d = 9, 120

= a + 28d = 9,120 – – – – – – – – – – – – – – – – – (2)

Subtract (2) from (1)

–18d = –4320

d = N240

Substituting d = N240 into (1)

a + 10(240) = 4800

The starting salary is N2400, while the annual increment is N N240

(ii) The annual salary at the time of retirement will be

T_n = a + (n – 1)d

T₃₆ = 2400 + (36 – 1) 240

T₃₆ = 2400 + 35 × 240

T₃₆ = N10800

Therefore, the monthly salary will be = N900

Example

A firm “SONY ELECTRONICS” produces 5000 sets of video CDs during its first year. The total sum of the firm production at the end of 10 years of production is 72,500set

(a) Estimate by how many units, production increased each year, if the increase each year is uniform

(b) Forecast, based on the estimate of annual increment in production, the level of the output for the 15th year.

Solution

(a) 1st year production = a = 5,000

Total production after 10 year = S₁₀ = 72,500

S₁₀ = [2 × 5000 + (10 – 1)d]

72,500 = 5 (10000 + 9d)

72,500 = 50000 + 45d

22500 = 45d

d = 500

500 units of Video CDs increase production each year.

(b) The production level in the 15th year

T₁₅ = 5000 + (15 – 1) 500

= 5000 + 7000

T₁₅ = 12,000

Sony Electronics will be producing 12,000 units of Video CDs in the

Example

Mr. Lagbaja took a loan of N200 from Mr. Tamedo and agrees to repay in number of installment, each installment (beginning with the second) exceeding the previous one by N10. If the first installment is N5. Find how many installments will be necessary to wipe the loan completely.

Solution

a = N5, d = N10, S_n = N2000

S_n = [2a +(n – 1)d]

2000 = [10 + (n – 1)10]

4000 = 10n + 10n² – 10n

4000 = 10n²

4000 = n²

n = + 20

n = 20 {n cannot be negative integer}

Mr Lagbaja will have to make 20 instalments

Example

Mr. Lawal borrows N1000 and at overall interest of N140. He repays the loan and interest by 12 monthly instalments each less by N10 than the preceding one. Find the amount of the first instalment.

Solution

Total money to be repaid = N1000 + N140 {Principal amount borrowed + interest}

= N1140

S₁₂ = 1140, n = 12, d = –10

S_n = [2a + (n – 1)d]

1140 = [2a + (12 – 1)(– 10)]

1140 = 6 (2a – 110)s

190 = 2a – 110

300 = 29

a = 150

The first instalment is N150

Example

Ejire venture produced 600 units of nylon pure water in the 3^rd year of its existence and 700 units in its 7^th year, what was the initial production in the first year?

Solution

T_n = a + (n – 1)d

T₃ = 600 = a + 2d

T₇ = 700 = a + 6d

Solving (i) and (ii) Simultaneously

d = 25, a = 550

The initial production in the first year is 550 unit of nylon pure water.

EXPONENTIAL SEQUENCE OR GEOMETRIC SEQUENCE

The sequence a, ar, ar², ar³ – – – ar^n–1 where a is the first term and any two consecutive number differs by a factor r (common ratio) is a Geometric Progression or Exponential Sequence.

T_n = ar^n–1

Consider the sequence

T₁, T₂, T₃, T₄, – – – – T_n–1, T_n which form a geometric progression

The common ratio is

Example

Write the first five terms of the geometric sequence whose first term is 6, with a common ratio of 3.

Solution

T₁ = a = 6

T₂ = ar = 6 x 3 = 18

T₃ = ar² = 6 x 3² = 54

T₄ = ar³ = 6 × 3³ = 162

T₅ = ar⁴ = 6 × 3⁴ = 486

Geometric sequence = 6, 18, 54, 162, 486 – – – –

Example

Find the 10th term of the geometric sequence given that a = 4, r = ½ and n = 10

Solution

T_n = ar^n–1

T₁₀ = 4 × (½)^10–1

= 4 × (½)⁹

Example

Find the 8th term of an geometric sequence whose first term is 3 whose common ratio is 2.

Solution

a = 3, r = 2, n = 8

T_n = ar^n–1

T₈ = 3 x 2^8–1

T₈ = 3 × 2⁷ = 3 × 128

T₈ = 384

Example

The 3^rd term of a geometric sequence is 27, while the 5^th term is 243, find the common ration, the first term and the 10^th term

Solution

T₃ = 27 = ar^3–1

27 = ar² – – – – – – – (i)

T₅ = 243 = ar^5–1

= 243 = ar⁴ – – – –– –(ii)

Divide (ii) by (i)

9 = r²

3² = r²

r = 3

Substitute r = 3 into (i)

27 = a × 3²

= 3

The first term is 3

The 10^th term will be

T₁₀ = 3 x 3^10–1

= 3 x 3⁹ = 3¹⁰

= 59049

GEOMETRIC MEAN

Consider A, B, C, are consecutive term of a geometric progression, then the common can be written

The term B that is the positive square root of the product A and C is called the geometric mean of A and C

Example

Find the geometric mean of 4 and 25

Solution

Let the geometric mean be B

Then 4, B, 25

example

Insert 4 geometric mean between 12 and 2916

Solution

Let the geometric means be A, B, C, D then 12, A, B, C, D, 2916 are in the geometric progression.

A = 12, n = 6, T₆ = 2916

T_n = ar^n–1

2916 = 12r⁵

243 = r⁵

3⁵ = r⁵

r = 3

T₂ = A = ar = 12 x 3 = 36

T₃ = B = ar² = 12 x 3² = 108

T₄ C = ar³ = 12 x 3³ = 324

T₅ = D = ar⁴ = 12 x 3⁴ = 972

The sequence is 12, 36, 108, 324, 972, 2916

SUM OF N TERM OF A GEOMETRIC PROGRESSION

S_n = a + ar + ar² + ar³ – – – – – – + ar^n–1 ––(1)

Multiply throughout by r

rS_n = ar + ar² + ar³ + – – – – – – – – – – – ar^n–1 + arⁿ –––(2)

Subtract (1) from (2)

rS_n – S_n = arⁿ – a

S_n (r – 1) = a(rⁿ–1)

The formula is applicable /r/</

Example

Find the sum of the first 10 terms of the geometric sequence, 3, 9, 27, 81, – –– – – – – – –

Solution

Example

Find the sum of the first five term of the G.P 128, 64, 32

Solution

Example

The Sum of the first two terms of a G.P is P and the sum of the last two term q. if the G.P has n terms. Find the common ratios.

Solution

Example

The first term of a geometric series is 3, the last term 768, if the sum of the term is 1533. Find the common ratio and the number of terms.

Solution

a = 3

l = ar^n–1 = 768

l = 3r^n–1 = 768

r^n–1 = 256 – – – – – – – – – – – – (1)

r² + 511r = –510 – – – – – (2)

From (1)

r^n–1 = 256

rⁿ = 256 r

rⁿ – 256r = 0 – – – – – – ––(3)

Subtract (3) from (2)

–255r = –510

Substituting r = 2 into (1)

2^n–1 = 256

2^n–1 = 2⁸

n – 1 = 8

n = 9

Example

The sum of the first n term of a geometric series of the first n term is 127 and the sum of their reciprocal is . The first term is 1. Find n and the common ratio

Solution

1 + r + r² + r³ + – – – – – – – + r^n–1 = 127 – – – – – – – – (1)

rⁿ – 1 = 127(r–1)

rⁿ – 127r = – 126 – – – – – – – – – – – – – – – – – – (3)

rⁿ = –126 + 127r

Sum of n term in (2)

From (3) rⁿ = 126 + 127r and substitute into (4)

127r^–1 – 64r^–n = 64

By inspection r = 2

When r = 2

127 = 2ⁿ – 1

128 = 2ⁿ

2⁷ = 2ⁿ

n = 7

The common ratio is 2, the number of term is 7

Example

Let the number be , a, ar

Their project , a, ar = 1728

a³ = 12³

a = 12

The sum of the numbers

38 = + 12 + 12r

38r = 12r² + 12r + 12

12r² – 26r + 12 = 0

6r² – 13r + 6 = 0

(2r – 3) (3r – 2) = 0

r = gives the number as 8, 12, 18

r = gives the number as 18, 12, 8

Hence the required numbers are 8, 12, 18

Example

The sum of the first six term of a G.P is 9 times the sum of first three terms. What is the common ratio.

Solution

Let the G.P be a, ar, ar² – – – – – –

1 – r⁶ = a(1 – r⁶)

(1 – r³)(1 + r³) = 9(1 – r³)

1 + r³ = 9

r³ = 8

= 2

INFINITE GEOMETRIC SERIES

If /r/</, we can evaluate the limiting value of the sum of an infinite ge3ometric series as n becomes larger and larger.

Let the limiting values of S_n = S, as n increases

Thus lim S_n = S

Example

Find the sum to infinite the geometric sequence

2,2(0.6), 2(0.6)², 2(0.6)³ – – – – – – 2(0.6)^n–1

Solution

a = 2, r = 0.6, /r/<1

Example

Find sum to infinity of the series

Sum to infinity of a G.P is

Solution

Example

Solution

Solution

Example

Solution

ARITHMETIC MEAN

Example

Solution

Example

Solution

Example:– If x2, y2, z2 are in AP, show that MPSetEqnAttrs('eq0011','',3,[[138,29,12,-1,-1],[184,37,16,-1,-1],[231,45,20,-1,-1],[209,42,19,-1,-1],[278,55,24,-1,-1],[347,68,30,-2,-2],[577,114,51,-3,-3]]); MPEquation();

MPSetEqnAttrs('eq0012','',3,[[276,174,77,-1,-1],[369,230,102,-1,-1],[461,287,128,-1,-1],[415,259,116,-1,-1],[554,345,154,-1,-1],[689,430,192,-2,-2],[1151,720,321,-3,-3]]); MPEquation();

Solution

SERIES

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SUM OF N TERMS OF NTH ARITHMETIC PROGRESSION

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Solution

APPLICATION OF ARITHMETIC PROGRESSION TO BUSINESS

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Solution

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Example

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Solution

EXPONENTIAL SEQUENCE OR GEOMETRIC SEQUENCE

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Solution

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Solution

GEOMETRIC MEAN

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SUM OF N TERM OF A GEOMETRIC PROGRESSION

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Solution

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Solution

Example

Example

Solution

INFINITE GEOMETRIC SERIES

Solution

Example:– If x², y², z² are in AP, show that