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.