## Geometric Sequence

The Geometric Sequence

Numbers are said to be in Geometric Sequence if there is a common ratio between any two consecutive terms.

Example:

In the sequence of the following numbers:

2, 4, 8, 16, 32, .....

The ratio between any two consecutive numbers is 2, i.e. 4/2 is same as 8/4.

In the above example of terms in geometric sequence, the common ratio is 2.

## Denotation of terms in a Geometric Sequence

The common ratio between any two consecutive terms is denoted as “r”.

Denote first term as “a”.

Then the terms in a geometric sequence can be denoted as

a, ar, ar2, ar3, .............arn – 1

## How to find the nth term or General term of a Geometric Sequence?

Recall that numbers are in Geometric Sequence if there is a common ratio between any two consecutive terms.

The common ratio is denoted by “r”

Let the terms of the geometric sequence be denoted as

a, ar, ar2, ar3, ............ , arn – 1

Then, the nth term or general term of a G.S. is

a r n – 1

Note:

The power in ‘r’ is one less than the nth term of a geometric sequence.
Therefore, 2nd term is ar, 3rd term is ar2, 4th term is ar3

Example:

1. The first term of a geometric sequence is 1/3. The common ratio is 3. Find the six terms of the G.S.

Solution:

First term, a = 1/3, common ratio, r = 3. Now, the other terms are:

Second term is ar = (1/3) × 3 = 1,

Third term is ar2 = (1/3) × 32 = 3,

Fourth term is ar3 = (1/3) × 33 = 9,

Fifth term is ar4 = (1/3) × 34 = 33 = 27,

Sixth term is ar5 = (1/3) × 35 = 34 = 81

2.  The fifth term of a geometric sequence is 625. The first term is unity. Find the 3rd term of the sequence.

Solution:

Recall that in a geometric sequence, the fifth term is denoted as ar4.

Now,

ar4 = 625, and a, the first term is 1, i.e. a = 1.

1 × r4 = 625, r4 = 625, r4 = 54, r = 4.

So, the third term denoted as ar2 is

1× 42 = 16

## How to find the sum of n terms of a geometric sequence?

Consider a geometric sequence with a common ratio ‘r’ and first term ‘a” in which the n terms are

a, ar, ar2, ar3, ....... arn-1

Let Sn denote the sum of the n terms of the above geometric sequence

Then,

Sn = a (rn – 1)/(r – 1), if r > 1, and

Sn = a (1 - rn)/ (1 - r), if r < 1

Example:

1. What is the sum of the first 10 terms of a geometric sequence in which the 5 terms are 2, 4, 8, 16, and 32?

Solution:

From the terms 2, 4, 8, 16, and 32, it can be seen that the

First term a = 2, and the common ratio, r = 2.

Since r > 1, apply the first of the above formulas to find sum of the 10 terms of the given geometric sequence

S10 = 2 (210 – 1)/ (2 – 1) = 2 (210 – 1)

Since, 210 is not a very large number, we will write its expansion:

210 = 1024

So, S10 = 2 (1024 – 1) = 2 × 1023 = 2046

2. What is the sum of the first 10 terms of a geometric sequence in which the first 3 terms are 1/3, 1/9 and 1/27?

Solution:

We have to find the sum given below:

1/3 + 1/9 + 1/27 +........... + 1/310

First term is a = 1/3, second term is 1/9,

Therefore common ratio is r = 1/9/1/3 = 1/3

Since r < 1 (as r = 1/3 < 1),

From the above formulas, apply the second one to find the sum of the first 10 terms of the given geometric sequence

Sn = a (1 - rn)/ (1 - r), if r < 1

S10 = [1/3 (1 – 1/310)]/ [(1 – 1/3)]

[1/3 (310 – 1)]/ [310 × 2/3] = (310 – 1)/2 × 310

Therefore, 1/3 + 1/9 + 1/27 +........... + 1/310 = (310 – 1)/2 × 310