Sum of GP to Infinity – Example and Proof

Here you will learn sum of gp to infinity (sum of infinite gp) and its proof with examples.

Let’s begin –

Sum of GP to Infinity (Sum of Infinite GP)

The sum of an infinite GP with first term a and common ratio r(-1 < r < 1 i.e. ,  | r | < 1) is

S = \(a\over 1-r\)

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

Proof : Consider an infinite GP with first term a and common ratio r, where -1 < r < 1 i.e. , | r | < 1. The sum of n terms of this GP is given by

\(S_n\) = a\({1-r^n}\over {1-r}\) = \(a\over {1-r}\) – \(ar^n\over {1-r}\)           ……..(i)

Since -1 < r < 1, therefore \(r^n\) decreases as n increases and \(r^n\) tends to zero as n tends to infinity i.e. \(r^n\) \(\rightarrow\) 0 as n \(\rightarrow\) \(\infty\).

\(\therefore\)    \(ar^n\over {1-r}\) \(\rightarrow\) 0 as n \(\rightarrow\) \(\infty\).

Hence from (i), the sum of an infinite GP is given by

S = \(lim_{n \to \infty}\) \(S_n\) = \(lim_{n \to \infty}\) ( \(a\over {1-r}\) – \(ar^n\over {1-r}\) ) = \(a\over 1-r\), if | r | < 1

Note : If r \(\ge\) 1, then the sum of an infinite GP tends to infinity.

Example : Find the sum to infinity of the GP \(-5\over 4\), \(5\over 16\), \(-5\over 64\), ……

Solution : The given GP has the first term a = -5/4 and the common ratio r = -1/4

Also | r | < 1.

Hence the sum of an infinite GP is given by S = \(a\over {1-r}\)

S = \(-5/4\over {1-(-1/4)}\) = -1

Example : The sum of an infinite GP is 57 and the sum of their cubes is 9747, find the GP.

Solution : Let a be the first term and r be the common ratio of the GP. Then

Sum = 57 \(\implies\) \(a\over 1-r\) = 57 …….(i)

Sum of the cubes = 9747

\(\implies\) \(a^3\) + \(a^3r^3\) + \(a^3r^6\) + ….. = 9747

\(\implies\) \(a^3\over {1 – r^3}\) = 9747 ……..(ii)

Dividing the cube of (i) by (ii), we get

\(a^3\over {(1-r)}^3\) . \((1-r^3)\over a^3\) = \({(57)}^3\over 9747\)

\(\implies\) \(1 – r^3\over {(1 – r)}^3\) = 19

= \(1+r+r^2\over {(1-r)}^2\)

= \(18r^2\) – 39r + 18 = 0

\(\implies\) (3r-2)(6r-9) = 0

\(\implies\) r = 2/3 or r = 3/2

Hence r = 2/3 [ \(\because\) r \(\ne\) 3/2, because -1 < r < 1 for an infinite GP]

Putting r = 2/3 in equation (i), we get

\(a\over {(1-(2/3))}\) = 57 \(\implies\) a = 19

Hence, the GP is 19, 38/3, 76/9, …….

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