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)]
= \(\frac{5}{9}\left[ 10\times \frac{\left( {{10}^{n}}-1 \right)}{\left( 10-1 \right)}-n \right]\),
= \(\frac{5}{9}\left[ \frac{10}{9}\left( {{10}^{n}}-1 \right)-n \right]\),
= \(\frac{5}{81}\left[ {{10}^{n+1}}-10-9n \right]\).