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 the same as 8/4.*

In the above example of terms in a 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 the 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/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:

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*

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:

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. *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*, the second term is *1/9, t*herefore the common ratio is *r = 1/9/1/3 = 1/3, s*ince 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*