# 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 = $$a\over 1-r$$

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, …….