# Formula for Geometric Progression (GP)

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$$({1\over r})^{n-1}$$

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.

Three positive integers form an increasing GP. If the middle term in this GP is doubled, then new numbers are in AP. Then the common ratio of GP is