 # Geometric Progression And Sum Of GP

Before going to learn how to find the sum of a given Geometric Progression, first know what is a GP in detail.

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.

$\frac{4}{2} = \frac{8}{4} = \frac{16}{8} = \frac{32}{16} = 2$

Similarly,

Consider a series $1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16},……….$

$\frac{\frac{1}{2}}{1}$ = $\frac{\frac{1}{4}}{\frac{1}{2}}$ = $\large \frac{\frac{1}{8}}{\frac{1}{4}} = \frac{\frac{1}{16}}{\frac{1}{8}} = \frac{1}{2}$

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

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

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

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

Also check: Arithmetic Progression

## Sum of Nth terms of G.P.

Consider the G.P,

$a,ar,ar^2,…..ar^{n-1}$

Let $S_n,a,r$ be the sum of n terms, first term and ratio of the G.P respectively.

Then, $S_n$ = $a + ar + ar^2 + ⋯ + ar^{n-1}$ —(1)

There are two cases, either $r = 1$ or $r ≠ 1$

If r=1, then $S_n$ = $a + a + a + ⋯ a$ = $na$

When $r ≠ 1$,

Multiply (1) with r gives,

$rS_n$ = $ar + ar^2 + ar^3 + ⋯ + ar^{n-1} + ar^n$—(2)

Subtracting (1) from (2) gives

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

$(r – 1) S_n = ar^{n} – a = a(r^{n}-1)$

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

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

### Solved Examples

Example 1: If $n^{th}$ 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

$r$= $\frac{6}{3}$ = $2$

Since the first term, $a$ = $3$

$a_n$ = $ar^{n-1}$

$192$ = $3 \times 2^{n-1}$

$2^{n-1} = \frac{192}{3} = 64 = 2^6$

$n – 1 = 6$

n = 7

Therefore, 192 is $7^{th}$ term of the G.P.

Example 2: $5^{th}$ term and $3^{rd}$ term of a G.P is 256 and 16 respectively. Find its $8^{th}$ term.

Solution: $ar^4$ = $256$—(1)

$ar^2$ = $16$—(2)

Dividing (1) by (2) gives,

$\frac{ar^4}{ar^2}$ = $\frac{256}{16}$

$r^2$ = $16$

$r$= $4$

Substituting $r$ = $4$ in (2) gives,

$a×4^2$ = $16$, $a$= $1$

$a_8$ = $ar^7$

=$1×4^7$ = $16384$

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

Solution: $a$ = $4$

Common ratio,$r = \frac{12}{4} = 3$

$n$ = $6$

Sum of n terms of a G.P,

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

$S_6$ = $\frac{4(3^6-1)}{(3-1)}$

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

1. teneshwari

how to solve cube root?

1. lavanya

Hi,