Sum of Infinite gp

The expansion G.P is geometric progression. It is a series where there is a common ratio between two consecutive terms. In a G.P if ‘a’ is the first term and ‘r’ is the common ratio, the nth term of the series can be calculated using the formula:


tn = arn-1.

Sum of the n terms can be calculated using the formula

(a (1 ** r ^n) )/ (1 ** r)

Where r < 1 or (a (r ^n **1)) / (1 ** r) if r > 1.

But sum to infinity of a G.P can be calculated by using the formula,

Sα = a / (1** r), (r <1 p="">
Here if r > 1, Sα = infinity.

Now let us see few problems on this topic sum of infinite G.P

Example Problems on Sum of Infinite Gp

Ex 1: Find the sum upto infinity of the following G.P: 2, 4, 8 …… upto infinity terms.

Soln: Here a = 2, r = 4 / 2 = 2

Therefore Sα = infinity as r > 1.

Ex 2: Find the sum upto infinity of the following G.P : 1 / 2, 1 / 4, 1 / 8  …… upto infinity term.

Soln: Here a = 1 / 2 , r = (1 / 4 )/ (1 / 2) = 1 / 2  < 1.

Therefore Sα = a / (1 ** r) = (1 / 2) / (1 **1 / 2)  = (1 / 2) / (1 / 2) = 1.

Therefore Sα = 1.

Ex 3: Find the sum of infinity of the G.P 1 / 3, 1 / 9, 1 / 27 ……..

Soln: Given: a = 1 / 3 , r = (1 / 9) / (1 / 3) = 1 / 3 < 1

Therefore Sα = a / (1 ** r) =( 1 / 3) / (1 ** 1 / 3 ) = (1 / 3) /( 2 / 3) = 1 / 2 .

Ex 4: The fourth term of a G.P is 2 / 27 and 7th term is 2 / 729 . Find the sum of infinity.

Sol: Given: T4 = 2 / 27 => ar 4-1 = 2 / 27

Therefore ar3 = 2 / 27 ………… (1)

Also, T7 = 2 / 729 => ar7 – 1 = 2 / 729

=> ar6 = 2 / 729 …………… (2)

Therefore (2) / (1) => ( ar ^6) /( ar ^3) = (2 / 729) /( 2 / 27)

=> r 3 = 27 / 729 = 1 / 27 = 1 / 3 ^3 .

Therefore (1) => a = 2 / 27 xx 1 / r ^3 = 2 / 27 xx (1 / 3) ^3 = (2 xx 27) / 27 = 2.

Therefore a = 2.

Therefore Sα = a / (1 ** r) = 2 / (1 ** 1 / 3) = 2 / (2 / 3) = 3.

Therefore Sα = 3.

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Practice Problems on Sum of Infinite Gp

1. 1 / 4, 1 / 16, 1 / 64 ……

[Ans: Sα = 1 / 3 ]

2. 2 / 3, 2 / 9, 2 / 27 ………..

[Ans: Sα = 1]