# Sum To n Terms Of a GP

A geometric progression (GP) can be written as a, ar, ar2, ar3, … arn – 1 in case of a finite GP and a, ar, ar2,…,arn – 1… in case of an infinite GP. We can calculate the sum to n terms of GP for finite and infinite GP using some formulas. Also, it is possible to derive the formula to find the sum of finite and finite GP separately. The sum to n terms of a GP refers to the sum of first n terms of a GP. In this article, you will learn how to derive the formula to find the sum of n terms of a given GP in different cases along with solved examples.

## Sum of First n Terms of GP Formula

The sum of first n terms of a GP a, ar, ar2, ar3,….arn-1,… is given be the formula,

Sn = a[(rn-1)/(r-1)] if r ≠ 1

Here,

a = First term

r = Common ratio

n = Number of terms

If the common ratio is equal to 1, then the sum of the first n term of the GP is given by:

Sn = na

Also, check: Geometric Sequence Calculator

## Sum of n terms of GP Proof

The proofs for the formulas of sum of the first n terms of a GP are given below.

Proof 1:

Suppose a, ar, ar2, ar3,….arn-1,… are the first n terms of a GP such that r ≠ 1.

Thus, a is the first term of r is the common ratio of the GP.

Let us denote by Sn the sum to first n terms of G.P.

So, Sn = a + ar + ar2 + ar3 + … + arn-1…..(i)

Multiplying (i) by r on both sides,

rSn = r(a + ar + ar2 + ar3 + … + arn-1)

rSn = ar + ar2 + ar3 + ar4 + ….+ arn….(ii)

Subtracting (i) from (ii),

rSn – Sn = (ar + ar2 + ar3 + ar4 + ….+ arn) – (a + ar + ar2 + ar3 + … + arn-1)

(r – 1)Sn = arn – a

(r – 1)Sn = a(rn – 1)

Sn = [a(rn – 1)]/ (r – 1)

Or

Sn = [a(1 – rn)]/ (1 – r)

Proof 2:

Suppose a, ar, ar2, ar3,….arn-1,… are the first n terms of a GP.

Here, a is the first term of r is the common ratio of the GP.

So, Sn = a + ar + ar2 + ar3 + … + arn-1…..(i)

If the common ratio is equal to 1, i.e. r = 1, then Sn can be written as:

Sn = a + a(1) + a(1)2 + a(1)3 + … + a(1)n-1

Sn = a + a + a + a + …. + a (n terms)

Sn = na

Thus, the above formulas can be written as:

The sum of first n term of a GP when r > 1

$$\large S_{n}=\frac{a(r^{n}-1)}{(r-1)}$$

The sum of first n term of a GP when r < 1

$$\large S_{n}=\frac{a(1-r^{n})}{(1-r)}$$

The sum of first n term of a GP when r = 1

$$\large S_{n}=na$$

The above formulas can be used to calculate the finite terms of a GP. Now, the question is how to find the sum of infinite GP. Let’s have a look at the formula given in the next section to know the formula of sum of infinite GP.

### Sum of Infinite GP

Consider an infinite GP, a, ar, ar2, ar3,…, arn-1, arn, …..

Here, a is the first term and r is the common ratio of the GP and the last term is not known.

Thus, the sum of infinite GP is given by the formula:

$$\large S_{n}=\frac{a}{(1-r)}$$

When |r| < 1, the geometric series converges and the series has a sum.

When |r| > 1 the geometric series does not converge or it is divergent and it has no sum.

### Solved Examples

Example 1:

Find the sum of first n terms of the GP: $$1 + \frac{2}{3}+\frac{4}{9}+….$$

Solution:

Given GP:

$$1 + \frac{2}{3}+\frac{4}{9}+….$$

Here,

First term = a = 1

Common ratio = r = 2/3, i.e. r < 1

Thus, the sum of first n terms is:

$$S_{n}=\frac{a(1-r^{n})}{(1-r)}$$

Substituting a = 1 and r = 2/3,

$$S_{n}=\frac{1(1-\frac{2}{3}^{n})}{(1-\frac{2}{3})}$$ $$S_{n}=\frac{3(1-\frac{2}{3}^{n})}{(3-2)}$$

Therefore,

$$S_{n}=3\left ( 1-\frac{2}{3}^{n} \right )$$

Example 2:

Find the sum of the first 6 terms of a GP whose first term is 2 and common difference is 4.

Solution:

Given,

First term = a = 2,

Common ratio = r = 4

and n = 6

As we know, the sum of first n terms of GP when r > 1

$$S_{n}=\frac{a(r^{n}-1)}{(r-1)}$$

Substituting the values,

$$S_{6}=\frac{2(4^{6}-1)}{(4-1)}$$

= (2/3) × (4096 – 1)

= (2/3) × 4095

= 2 × 1365

= 2730

Therefore, the sum of the first 6 terms of the given GP is 2730.

Example 3:

Find the sum of the infinite GP: 125, 25, 5,….

Solution:

Given GP is,

125, 25, 5,….

Here,

First term = a = 125

Common ratio = r = 25/125 = 1/5; r < 1

The sum of an infinite GP when r < 1 is:

Sn = a/(1 – r)

= 125/[1 – (1/5)]

= 125/[(5 – 1)/5]

= (125 × 5)/4

= 625/4

= 156.25

Hence, the sum of given infinite GP is 156.25.

### Practice Problems

1. Which term of the GP, 2, 8, 32, … up to n terms is equal to 131072?

2. Find the sum of the sequence 7, 77, 777, 7777, … to n terms.

3. How many terms of G.P. 3, 32, 33,… are needed to give the sum 120?

4. If the sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2, respectively, then find the last term and the number of terms.

## Frequently Asked Questions – FAQs

### What is the formula of sum of n terms in GP?

The formula of sum of n terms in GP is given as:
S_n = [a(r^n – 1)]/(r – 1) when |r| > 1
S_n = [a(1 – r^n)]/(1 – r) when |r|

### What is the nth term of GP?

The nth term of a GP is denoted by a_n and is calculated using the formula:
a_n = ar^{n-1}
Here, a is the first term and r is the common ratio of the GP.

### What is the sum of infinite GP?

The sum of infinite GP whose first term a and common ratio r is given by the formula,
S_{n} = a/(1 – r)

### Is it possible to find an infinite sum in a GP?

Yes, we can find the sum of an infinite GP only when the common ratio is less than 1. If the common ratio is greater than 1, there will be no specified sum as we can say that the sum is infinity.

### What is the sum of first n terms of a GP when the common ratio is one?

If the common ratio is equal to 1, then the GP becomes, a, a, a,…
Thus, the sum of first n terms = a + a + a + … (n terms) = na