 # Geometric Progression And Sum Of GP

In geometric progression (G.P.), the sequence is geometric and is a result of the sum of G.P. A geometric series is the sum of all the terms of geometric sequence. Before going to learn how to find the sum of a given Geometric Progression, first know what a GP is in detail.

## What is Geometric Progression?

If in a sequence of terms, each succeeding term is generated by multiplying each preceding term with a constant value, then the sequence is called a geometric progression. (GP), whereas the constant value is called the common ratio. For example, 2, 4, 8, 16, 32, 64, … is a GP, where the common ratio is 2.

$$\begin{array}{l}\frac{4}{2} = \frac{8}{4} = \frac{16}{8} = \frac{32}{16} = 2\end{array}$$

Similarly,

Consider a series
$$\begin{array}{l}1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16},……….\end{array}$$
$$\begin{array}{l}\frac{\frac{1}{2}}{1}\end{array}$$
=
$$\begin{array}{l}\frac{\frac{1}{4}}{\frac{1}{2}}\end{array}$$
=
$$\begin{array}{l}\large \frac{\frac{1}{8}}{\frac{1}{4}} = \frac{\frac{1}{16}}{\frac{1}{8}} = \frac{1}{2}\end{array}$$

In the given examples, the ratio is a constant. Such sequences are called Geometric Progressions. It is abbreviated as G.P.

A sequence

$$\begin{array}{l}a_1,a_2,a_3,…….a_n,….\end{array}$$
is a G.P, then
$$\begin{array}{l}\frac{a_{k+1}}{a_k} = r~~~~~~~~~~~~ [k>1]\end{array}$$

Where r is a constant which is known as common ratio and none of the terms in the sequence is zero.

Now, learn how to add GP if there are n number of terms present in it.

Also check: Arithmetic Progression

## Sum of Nth terms of G.P.

A geometric series is a sum of an infinite number of terms such that the ratio between successive terms is constant. In this section, we will learn to find the sum of geometric series.

### Derivation of Sum of GP

Since, we know, in a G.P., the common ratio between the successive terms is constant, so we will consider a geometric series of finite terms to derive the formula to find the sum of Geometric Progression.

Consider the G.P,

$$\begin{array}{l}a,ar,ar^2,…..ar^{n-1}\end{array}$$
Let
$$\begin{array}{l}S_n,a,r\end{array}$$
be the sum of n terms, first term and ratio of the G.P respectively.
Then,
$$\begin{array}{l}S_n\end{array}$$
=
$$\begin{array}{l}a + ar + ar^2 + ⋯ + ar^{n-1}\end{array}$$
—(1)

There are two cases, either

$$\begin{array}{l}r = 1\end{array}$$
or
$$\begin{array}{l}r ≠ 1\end{array}$$

If r=1, then

$$\begin{array}{l}S_n\end{array}$$
=
$$\begin{array}{l} a + a + a + ⋯ a\end{array}$$
=
$$\begin{array}{l}na\end{array}$$

When

$$\begin{array}{l}r ≠ 1\end{array}$$
,

Multiply (1) with r gives,

$$\begin{array}{l}rS_n\end{array}$$
=
$$\begin{array}{l} ar + ar^2 + ar^3 + ⋯ + ar^{n-1} + ar^n\end{array}$$
—(2)

Subtracting (1) from (2) gives

$$\begin{array}{l}rS_n – S_n = (ar + ar^{2} + ar^{3} + …. + ar^{n-2} + ar^{n-1} + ar^{n}) – (a + ar + ar^{2} + …. + ar^{n-2} + ar^{n-1})\end{array}$$

$$\begin{array}{l}(r – 1) S_n = ar^{n} – a = a(r^{n}-1)\end{array}$$

$$\begin{array}{l}S_n = a\frac{(r^{n}-1)}{(r – 1)} = a\frac{(1 – r^{n})}{(1 – r )}\end{array}$$
 Sum of n terms  Sn = $$\begin{array}{l}\frac{a(r^n – 1)}{r-1}\end{array}$$; Where r $$\begin{array}{l}\neq\end{array}$$ 1

The above formula is also called Geometric Progression formula or G.P. formula to find the sum of GP of finite terms. Here, r is the common ratio of G.P. formula.

## Sum of GP for Infinite Terms

If the number of terms in a GP is not finite, then the GP is called infinite GP. The formula to find the sum to infinity of the given GP is:

$$\begin{array}{l}S_{\infty}=\sum_{n=1}^{\infty} ar^{n-1}=\frac{a}{1-r}; -1<r<1\end{array}$$

Here,

S = Sum of infinite geometric progression

a = First term of G.P.

r = Common ratio of G.P.

n = Number of terms

This formula helps in converting a recurring decimal to the equivalent fraction. This can be observed from the following example:

0.22222222… = 0.2 + 0.02 + 0.002 + 0.0002 + …..∞

= (0.2 × 0.10) + (0.2 × 0.11) + (0.2 × 0.12) + (0.2 × 0.13) + ….∞

= (0.2) + (0.2 × 0.1) + (0.2 × 0.12) + (0.2 × 0.13) + ….∞

This of the form a + ar + ar2 + ar3 + ….. ∞ (infinite GP) such that a = 0.2 and r = 0.1.

Thus, 0.22222222… = 0.2/(1 – 0.1)

= 0.2/0.9

= 2/9

Hence, the equivalent fraction of 0.22222222… is 2/9.

## Video Lesson

### Sum of Infinite Terms of G.P. ## Solved Examples on Sum of G.P.

Example 1: If
$$\begin{array}{l}n^{th}\end{array}$$
term of the G.P 3, 6, 12, …. is 192, then what is the value of n?

Solution: First, we have to find the common ratio

$$\begin{array}{l}r\end{array}$$
=
$$\begin{array}{l}\frac{6}{3}\end{array}$$
=
$$\begin{array}{l}2\end{array}$$

Since the first term,

$$\begin{array}{l}a\end{array}$$
=
$$\begin{array}{l}3\end{array}$$

$$\begin{array}{l}a_n\end{array}$$
=
$$\begin{array}{l}ar^{n-1}\end{array}$$

$$\begin{array}{l}192\end{array}$$
=
$$\begin{array}{l}3 \times 2^{n-1}\end{array}$$

$$\begin{array}{l}2^{n-1} = \frac{192}{3} = 64 = 2^6\end{array}$$

$$\begin{array}{l}n – 1 = 6 \end{array}$$

n = 7

Therefore, 192 is

$$\begin{array}{l}7^{th}\end{array}$$
term of the G.P.
Example 2:
$$\begin{array}{l}5^{th}\end{array}$$
term and
$$\begin{array}{l}3^{rd}\end{array}$$
term of a G.P is 256 and 16 respectively. Find its
$$\begin{array}{l}8^{th}\end{array}$$
term.

Solution:

$$\begin{array}{l}ar^4\end{array}$$
=
$$\begin{array}{l}256\end{array}$$
—(1)

$$\begin{array}{l}ar^2\end{array}$$
=
$$\begin{array}{l}16\end{array}$$
—(2)

Dividing (1) by (2) gives,

$$\begin{array}{l}\frac{ar^4}{ar^2}\end{array}$$
=
$$\begin{array}{l}\frac{256}{16}\end{array}$$

$$\begin{array}{l}r^2\end{array}$$
=
$$\begin{array}{l}16\end{array}$$

$$\begin{array}{l}r\end{array}$$
=
$$\begin{array}{l}4\end{array}$$

Substituting

$$\begin{array}{l}r\end{array}$$
=
$$\begin{array}{l}4\end{array}$$
in (2) gives,

$$\begin{array}{l}a×4^2\end{array}$$
=
$$\begin{array}{l}16\end{array}$$
,
$$\begin{array}{l}a\end{array}$$
=
$$\begin{array}{l}1\end{array}$$

$$\begin{array}{l}a_8\end{array}$$
=
$$\begin{array}{l}ar^7\end{array}$$

=

$$\begin{array}{l} 1×4^7\end{array}$$
=
$$\begin{array}{l}16384\end{array}$$

Example 3: Find the sum of the first 6 terms of the G.P 4, 12, 36,…..

Solution:

$$\begin{array}{l}a\end{array}$$
=
$$\begin{array}{l}4\end{array}$$

Common ratio,

$$\begin{array}{l}r = \frac{12}{4} = 3\end{array}$$

$$\begin{array}{l}n\end{array}$$
=
$$\begin{array}{l}6\end{array}$$

Sum of n terms of a G.P,

$$\begin{array}{l}S_n\end{array}$$
=
$$\begin{array}{l}\frac{a(r^n-1)}{(r-1)}\end{array}$$

$$\begin{array}{l}S_6\end{array}$$
=
$$\begin{array}{l}\frac{4(3^6-1)}{(3-1)}\end{array}$$

=

$$\begin{array}{l}\frac{4(729-1)}{(2)}\end{array}$$
=
$$\begin{array}{l}2 × 728 \end{array}$$
=
$$\begin{array}{l}1456\end{array}$$

## Frequently Asked Questions on Sum of Geometric Progression

### What is sum of geometric series?

The sum of a geometric series depends on the number of terms in it. The sum of a geometric series will be a definite value if the ratio’s absolute value is less than 1. If the numbers are approaching zero, they become insignificantly small. In this case, the sum to be calculated despite the series comprising infinite terms.

### How do you find the sum of a geometric series?

The formula used for calculating the sum of a geometric series with n terms is Sn = a(1 – r^n)/(1 – r), where r ≠ 1.

### How do you calculate GP common ratio?

To calculate the common ratio of a GP, divide the second term of the sequence with the first term or simply find the ratio of any two consecutive terms by taking the previous term in the denominator.

### What is r in GP?

In GP, the letter r represents the common ratio, which is the same for any two consecutive terms of the given GP.

### What is the sum to infinite GP?

The sum to infinite GP means, the sum of terms in an infinite GP. The formula to find the sum of infinite geometric progression is S_∞ = a/(1 – r), where a is the first term and r is the common ratio.

Test your knowledge on Geometric Progression Sum Of Gp

1. teneshwari

how to solve cube root?

1. lavanya

Hi,