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 an 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 an 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}}

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