# Geometric Series

The series obtained by adding the terms of a G.P. is called the geometric series. Let ${S_n}$ be the sum of the first $n$ terms of the G.P. The $nth$ term is ${a_1}{r^{n - 1}}$, the $\left( {n - 1} \right)th$ term is ${a_1}{r^{n - 2}}$, etc. Hence
${S_n} = {a_1} + {a_1}r + {a_1}{r^2} + {a_1}{r^3} + \cdots + {a_1}{r^{n - 1}}$   --- (1)

Multiplying both members of (1) by $r$, we get
$r{S_n} = {a_1}r + {a_1}{r^2} + {a_1}{r^3} + {a_1}{r^4} + \cdots + {a_1}{r^n}$ --- (2)

On subtracting each side of (2) from the corresponding side of (1), we obtain
${S_n} - r{S_n} = {a_1} - {a_1}{r^n}$
${S_n}\left( {1 - r} \right) = {a_1} - {a_1}{r^n}$
$\therefore$     ${S_n} = \frac{{{a_1}\left( {1 - {r^n}} \right)}}{{1 - r}}{\text{if }}r < 1$

If the common ratio in a G.P. is more than $1$, then each successive term is greater than the previous one and the sum of the terms grows very rapidly and tends to infinity as $n$ tends to infinity.

If the common ratio is less than $1$, then each term will be smaller than the previous one and the total will increase but will not exceed a finite value as $n$ tends to infinity.

Thus

Example:
Using G.P., find the sum of $1,2,4,8,16,32,64,128$.

Solution:
We have
${a_1} = 1$, $r = \frac{2}{1} = 2$ and $n = 8$

Substituting these values in the given formula, we get
${S_n} = \frac{{{a_1}\left( {{r^n} - 1} \right)}}{{r - 1}}$       (as $r > 1$)
${S_8} = \frac{{1\left( {{2^8} - 1} \right)}}{{2 - 1}} = 225$

Example:
Given $r = \frac{2}{3}$, $n = 6$, ${S_n} = \frac{{665}}{{144}}$, find ${a_1}$.

Solution:
Since, ${S_n} = \frac{{{a_1}\left( {1 - {r^n}} \right)}}{{1 - r}}$
$\frac{{665}}{{144}} = {a_1}\left[ {\frac{{1 - {{\left( {\frac{2}{3}} \right)}^6}}}{{1 - \frac{2}{3}}}} \right] = {a_1}\left[ {\frac{{1 - \frac{{64}}{{729}}}}{{\frac{1}{3}}}} \right]$
$\Rightarrow {a_1} = \frac{{27}}{{16}}$