Here you will learn what is geometric progression and formula for geometric progression with examples.

Let’s begin –

## Geometric Progression (GP)

A sequence of non-zero numbers is called a geometric progression (GP) if the ratio of a term and the term preceding to its always a constant quantity.

The constant ratio is called the common ration of the GP.

In other words, a sequence \(a_1\), \(a_2\), \(a_3\) …….. \(a_n\) , …….. is called a geometric progression if

\(a_{n+1}\over a_n\) = constant for all n \(\in\) N

Example : The sequence 4, 12, 36, 108, is a GP , because

\(12\over 4\) = \(36\over 12\) = \(108\over 36\) = ….. = 3, which is a constant.

Clearly, this sequence is a GP with first term 4 and common ratio 3.

**Also Read** : Formula for Arithmetic Progression (AP)

## Formula for Geometric Progression

### (a) General Term of a GP

Let a be the first term and r be the common ratio of a GP. Then its nth term or general term is \(ar^{n-1}\)

i.e \(a_n\) = \(ar^{n-1}\)

**Note : **The whole GP can be written as a, ar, \(ar^2\), …… , \(ar^{n-1}\) or a, ar, \(ar^2\), …… , \(ar^{n-1}\), …… according as it is finite or infinite.

### (b) nth term from the end of a finite GP

(i) The nth term from the end of a finite GP consisting of m terms is \(ar^{m-n}\), where a is the first term and r is the common ratio of the GP.

\(\because\) nth term from the end = (m – n + 1)th term from the beginning

= \(ar^{m-n}\)

(ii) and the nth term from the end of a GP with last term l and common ratio r is given by

\(a_n\) = l\(({1\over r})^{n-1}\)

Clearly when we look at the terms terms of a GP from the last term and move towards the beginning we find that the progression is a GP with the common ration 1/r.

So, nth term from the end = l\(({1\over r})^{n-1}\)

**Also Read** : Sum of GP Series Formula | Properties of GP

Example : Find the 9th term and the general term of the progression \(1\over 4\), \(-1\over 2\), 1, -2, ……

Solution : The given progresion is clearly a gp with the first term a = \(1\over 4\) and common ratio r = -2.

\(\therefore\) 9th term = \(a_9\) = \(ar^{(9-1)}\) = \(1\over 4\)\((-2)^{8}\) = 64

and, General term = \(a_n\) = \(ar^{(n-1)}\) = \(1\over 4\) \((-2)^{n-1}\) = \((-1)^{n-1}\) \(2{n-3}\)

### Related Questions

If the 4th and 9th terms of a G.P. be 54 and 13122 respectively, find the G.P.