 # Geometric Progression And Sum Of GP

## Geometric progression – Introduction:

If in a sequence of terms each term is constant multiple of the preceding term, then the  sequence is called a geometric progression.(G.P), whereas the constant multiplier is called the common ratio.

In an A.P, the difference between $n^{th}$ term and $(n-1)^{th}$ term will be a constant which is known as the common difference of the A.P.

An Arithmetic Progression is in the form,

$a, a+d, a+2d, a+3d ….. a + (n-1)d$

Now, what if the ratio of $n^{th}$ term to $(n-1)^{th}$ term in a sequence is a constant?

For example, consider the following sequence,

$2, 4, 8, 16, 32,……..$

You can see that,

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

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.

## General term of a Geometric Progression:

We had learned to find the $n^{th}$ term of an A.P, which was

$a_n$= $a + (n-1)d$

Similarly, in case of G.P,

Let a be the first term and r be the common ratio,

Then the second term, $a_2 = a \times r = ar$

Third term, $a_{3} = a_{2} \times r = ar \times r = ar^{2}$

Similarly, $n^{th}$ term, $a_{n}$ = $ar^{n-1}$

 General term of a Geometric Progression  an = arn−1

## Common Term

Consider the sequence a, ar, ar2, ar3,……

First term = a

Second term = ar

Third term = ar2

Similarly nth term, tn =  arn-1

Thus, Common ratio = (Any term) / (Preceding term)

= tn / tn-1

= (arn – 1 ) /(arn – 2)

= r

Thus, the general term of a G.P is given by  arn-1 and the general form of a G.P is a + ar + ar2  + …..

For Example: r = t2 / t1 = ar / a = r

## Finite and Infinite Geometric Progression

The terms of a finite G.P. can be written as $a, ar, ar^2, ar^3,……ar^{n-1}$

Terms of an infinite G.P. can be written as $a, ar, ar^{2}, ar^{3}, ……ar^{n-1},…….$

$a + ar + ar^2 + ar^3 + ⋯ + ar^n$ is called finite geometric series.

$a + ar + ar^2 + ar^3 + ⋯ + ar^n + ⋯$ is called infinite geometric series.

## Sum of n terms of a 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

### Example Problems

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 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,

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

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

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