# Infinite Geometric Series

## Infinite Geometric Series

Sum of n terms of a G.P: The sum of n terms of a G.P. with first term a and common ratio r is given by $${{S}_{n}}=a\left[ \frac{{{r}^{n}}-1}{r-1} \right]$$ or $${{S}_{n}}=a\left( \frac{1-{{r}^{n}}}{1-r} \right)\,\,\,\,r\ne 1$$.

Proof: Let Sn denote the sum of n terms of the G.P. with first term a and common ratio r

Then, Sn = a + ar + ar² + … + arn … (1)

Multiplying both sides by r, we get

rSn = ar + ar² + ar³ + … +arn … (2)

On substracting (2) from (1), we get

Sn – r Sn = a – arn

Sn (1 – r) = a (1 – rⁿ)

$${{S}_{n}}=a\left[ \frac{1-{{r}^{n}}}{1-r} \right]$$ or $${{S}_{n}}=a\left[ \frac{{{r}^{n}}-1}{r-1} \right]$$

If l is the last term of the G.P., then l = arⁿ⁻1

$${{S}_{n}}=a\left[ \frac{1-{{r}^{n}}}{1-r} \right]=\frac{a-a{{r}^{n}}}{1-r}=\frac{a-\left( a{{r}^{n-1}} \right)r}{1-r}=\frac{a-lr}{1-r}$$

Thus $${{S}_{n}}=\frac{a-lr}{1-r}$$ or $${{S}_{n}}=\frac{lr-a}{r-1},r\ne 1$$.

If |r| < 1 then $$\underset{n\to \infty }{\mathop{\lim }}\,{{r}^{n}}=0$$.

∴ The sum S of an infinite G.P. with common ratio r satisfying |r| < 1 is given by $$S=\underset{n\to \infty }{\mathop{\lim }}\,{{S}_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,a\left[ \frac{1-{{r}^{n}}}{1-r} \right]$$,

⇒ $$S=\frac{a}{1-r}$$.

Thus, the sum S of an infinite G.P. with first term a and common ratio r (-1 < r < 1) is given by $$S=\frac{a}{1-r}$$.Example: Find the sum of 5 + 55 + 555 + … to n terms

Solution: We have,

5 + 55 + 555 + … to n terms

= 5 [1 + 11 + 111 + … + to n terms]

= 5/9 [9 + 99 + 999 + … + to n terms]

= 5/9 [(10 – 1) + (10² – 1) + (10³ – 1) + … + to n terms]

= 5/9 [(10) + (10²) + (10³)) – (1 + 1 + 1 + … + 1)]

= $$\frac{5}{9}\left[ 10\times \frac{\left( {{10}^{n}}-1 \right)}{\left( 10-1 \right)}-n \right]$$,

= $$\frac{5}{9}\left[ \frac{10}{9}\left( {{10}^{n}}-1 \right)-n \right]$$,

= $$\frac{5}{81}\left[ {{10}^{n+1}}-10-9n \right]$$.