GEOMETRIC SEQUENCE AND SERIES

Geometric Sequence

A geometric sequence is formed by multiplying a non zero real number to get to the next term.

In other words, a geometric sequence is a sequence in which the ratio between any two consecutive terms is same along the sequence.

For example :

3, 6, 12, 24, 48, ....... has a common ratio of 2

80, 40, 20, 10, 5, ....... has a common ratio of 1/2

2, -6, 18, -54, 162, ....... has a common ratio of -3

In general, if the common ratio is r (which can be negative), then

1^st term : a₁

2^nd term : a₁r

3^rd term : a₁r²

4^th term : a₁r³

5^th term : a₁r⁴

From this, we can suggest a formula for the n^th term.

For a geometric sequence with common ratio r,

a_n = a₁r^{n - 1}

Example 1 :

A geometric sequence has first term 3 and common ratio -2. Find the 10^th term.

Solution :

Formula for n^th term of a geometric sequence :

a_n = a₁r^{n - 1}

Substitute n = 10, a₁ = 3 and d = -2.

a₁₀ = 3(-2)^{10 - 1}

= 3(-2)⁹

= 3(-512)

= -1536

Example 2 :

The second term of a geometric sequence is 6 and the fourth term is 96. Find the possible values of the first term and the common ratio.

Solution :

a₂ = 6 and a₄ = 96

a₂ = 6

a₁r^{2 - 1} = 6

a₁r = 6 ----(1)

a₄ = 96

a₁r^{4 - 1} = 96

a₁r³ = 96 ----(2)

(2) ÷ (1) :

r² = 16

Take square root on both sides.

r = ±4

The possible values of the common ratio are -4 and 4.

Substitute r = -4 in (1).

a₁(-4) = 6

a₁ = -3/2

Substitute r = 4 in (1).

a₁(4) = 6

a₁ = 3/2

The possible values of the first term are -3/2 and 3/2.

Example 3 :

Find the number of terms in the geometric sequence :

1, 2, 4, 8, ......., 512

Solution :

This is a geometric sequence with the first term 1 and common ratio 2.

Let a_n = 512.

a₁r^{n - 1}= 512

Substitute a₁ = 1 and r = 2.

1(2)^{n - 1}= 512

Write 512 as a power of 2.

2^{n - 1}= 2⁹

n - 1 = 9

n = 10

Geometric Series

Just as for arithmetic sequences, there is a formula for finding the sum of the first n terms of a geometric sequence. This sum is sometimes called a geometric series.

For a geometric sequence with common ratio r ≠ 1 :

S_n = a₁(1 - rⁿ)/(1 - r)

S_n = a₁(rⁿ- 1)/(r - 1)

If r = 1,

S_n = na₁

Proof

Prove that for a geometric sequence with common ratio r ≠ 1,

S_n = a₁(1 - rⁿ)/(1 - r)

Write out the first few terms and the last few terms.

S_n= a₁ + a₁r + a₁r² + .......... + a₁r^{n - 2} + a₁r^{n - 1}----(1)

Multiply each side by r.

rS_n= a₁r + a₁r² + a₁r³+......... + a₁r^{n - 1} + a₁rⁿ----(2)

(1) - (2) :

S_n- rS_n= a₁- a₁rⁿ

S_n- rS_n= a₁(1 - rⁿ)

S_n(1 - r)= a₁(1 - rⁿ)

S_n= a₁(1 - rⁿ)/(1 - r)

For a geometric sequence with common ratio r = 1 :

S_n = na₁

Proof

Prove that for a geometric sequence with common ratio r = 1,

S_n = na₁

Write out the first few terms and the last few terms.

S_n= a₁ + a₁r + a₁r² + .......... + a₁r^{n - 2} + a₁r^{n - 1}

Substitute r = 1.

S_n= a₁ + a₁(1) + a₁(1)² + .......... + a₁(1)^{n - 2} + a₁(1)^{n - 1}

S_n= a₁ + a₁ + a₁ + .......... + a₁ + a₁(to n terms)

S_n= na₁

Sum to Infinite Terms of a Geometric Sequence

The formula given below can be used to find the sum to infinite terms of a geometric sequence.

S_∞ = a₁/(1 - r), -1 < r < 1

If the value of r does not lie between -1 and 1, then the sum to infinite terms of a geometric sequence can not be found.

Example 4 :

Find the sum of 10 terms of the geometric sequence :

1, 2, 4, 8, ..........

Solution :

This is a geometric sequence with a₁ = 1 and r = 2.

Formula for the sum of first n terms of a geometric sequence.

S_n = a₁(1 - rⁿ)/(1 - r)

Substitute n = 10, a₁ = 1 and r = 2.

S₁₀ = 1(1 - 2¹⁰)/(1 - 2)

= 1(1 - 1024)/(-1)

= -1023/(-1)

= 1023

Example 5 :

Find the sum of first 8 terms of a geometric sequence whose n^th term 3^2n-1.

Solution :

a_n = 3^2n-1

a₁ = 3^{2(1) - 1}

= 3^{2 - 1}

= 3¹

= 3

a₂ = 3^2(2)-1

= 3^{4 - 1}

= 3³

= 27

Common ratio :

r = a₂/a₁

r = 27/3

r = 9

Formula for the sum of first n terms of a geometric sequence.

S_n = a₁(1 - rⁿ)/(1 - r)

Substitute n = 8, a₁ = 3 and r = 9.

S₈ = 3(1 - 3⁸)/(1 - 3)

= 3(1 - 6561)/(-2)

= 3(-6560)/(-2)

= 3(3280)

= 9840

Example 6 :

Find the first term of a geometric sequence whose common ratio is 5 and sum to first 6 terms is 46872.

Solution :

S₆ = 46872

a₁(1 - r⁶)/(1 - r) = 46872

Substitute r = 5.

a₁(1 - 5⁶)/(1 - 5) = 46872

a₁(1 - 15625)/(-4) = 46872

a₁(-15624)/(-4) = 46872

3906a₁ = 46872

Divide each side by 3906.

a₁ = 12

Example 7 :

Find the sum of the geometric series :

2 + 6 + 18 + ........ + 13122

Solution :

This is a geometric series with a₁ = 2 and r = 3.

Let a_n= 13122.

a_n = 13122

a₁r^{n - 1} = 13122

Substitute a₁= 2 and r = 3.

2(3)^{n - 1} = 13122

3^{n - 1} = 6561

Write 6561 as a power of 3.

3^{n - 1} = 3⁸

n - 1 = 8

n = 9

Formula for the sum of first n terms of a geometric sequence.

S_n = a₁(1 - rⁿ)/(1 - r)

Substitute n = 9, a₁ = 2 and r = 3.

S₉ = 2(1 - 3⁹)/(1 - 3)

= 2(1 - 19683)/(-2)

= 2(-19682)/(-2)

= 19682

Example 8 :

Find the sum of the geometric series :

5 + 5 + 5 + ........ to 27 terms

Solution :

This is a geometric series with a₁ = 5 and r = 1.

S_n = na₁

Substitute n = 27 and a₁= 5.

S_n = 27(5)

= 135

Example 9 :

Find the sum to infinity of 9 + 3 + 1 + ........

Solution :

This is a geometric series with a₁ = 9 and r = 1/3.

S_∞= a₁/(1 - r)

Substitute a₁= 9 and r = 1/3.

S_∞= 9/(1 - 1/3)

= 9/(2/3)

= 9 ⋅ 3/2

= 27/2

Example 10 :

Peterson writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with the instruction that they continue the process similarly. Assuming that the process is unaltered and it costs $2 to mail one letter, find the amount spent on postage when 8^th set of letters is mailed.

Solution :

Amount spent when the first set of letters is mailed :

= 4 ⋅ 2

= $8

Amount spent when the second set of letters is mailed :

= 4 ⋅ 4 ⋅ 2

= $32

Amount spent when the third set of letters is mailed :

= 4 ⋅ 4 ⋅ 4 ⋅ 2

= $128

If this pattern continues, we will have a geometric sequence with the first term 8 and common ratio 4 as shown below.

8, 32, 128, ............. to 8 terms

Find the sum of the terms in the above geometric sequence.

S_n = a₁(1 - rⁿ)/(1 - r)

Substitute n = 8, a₁ = 8 and r = 4.

S₈ = 8(1 - 4⁸)/(1 - 4)

= 8(1 - 65536)/(-3)

= 8(-65535)/(-3)

= 8(21845)

= $174,760

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GEOMETRIC SEQUENCE AND SERIES