# 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

\[{S_n} = \frac{{{a_1}\left( {1 – {r^n}} \right)}}{{1 – r}}{\text{ if }}r < 1{\text{}}\]

\[{S_n} = \frac{{{a_1}\left( {{r^n} – 1} \right)}}{{r – 1}}{\text{ if }}r > 1\]

** 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, $${S_n} = \frac{{{a_1}\left( {1 – {r^n}} \right)}}{{1 – r}}$$

$$\frac{{665}}{{144}} = {a_1}\left[ {\frac{{1 – {{\left( {\frac{2}{3}} \right)}^6}}}{{1 – \frac{2}{3}}}} \right] = {a_1}\left[ {\frac{{1 – \frac{{64}}{{729}}}}{{\frac{1}{3}}}} \right]$$

$$ \Rightarrow {a_1} = \frac{{27}}{{16}}$$