Let's consider the first term of a GP series is "a" and the common difference is "d". If the sum of n terms of a GP series is "S" then
\(S = a + a \times d + a \times d^2 +..\) \(..+ a \times d^{n - 1}....(1)\)
after multiplying by "d" on both sides of the equation (1)
\(S \times d = a \times d + a \times d^2 + a \times d^3 +..\) \(..+ a \times d^n....(2)\)
By subtracting equation (2) from equation (1)
\((S - Sd) = a - ad + ad - ad^2 + ad^2\) \( - ad^3.......+ ad^{n - 1} - ad^n\) $$ S \ (1 - d) = a \ (1 - d^n) $$ $$ S = \frac{a \ (1 - d^n)}{1 - d}.....(3) $$ It can also be written as $$ S = \frac{a \ (d^n - 1)}{d - 1} $$ If "l" is the last term of the series then $$ l = ad^{n - 1} $$ $$ l = \frac{ad^n}{d} $$ $$ ld = ad^n $$ By putting the value of \(ad^n\) in equation (3) $$ S = \frac{a \ (1 - d^n)}{1 - d} $$ $$ S = \frac{a - ad^n}{1 - d} $$ $$ S = \frac{a - ld}{1 - d}.....(4) $$
Ex(1): Find the sum of the given GP series till nine terms? $$ 1 + 2 + 4 + 8 + 16.... $$
Solution: From the given series
the first term of the series (a) = 1
Common difference (d) = 2
and n = 9, then $$ S_n = \frac{a \ (d^n - 1)}{d - 1} $$ $$ S_9 = \frac{1 \ (2^9 - 1)}{2 - 1} $$ $$ S_9 = \frac{512 - 1}{1} = 511 $$ Hence the sum of 9 terms of the given GP series is 511.
Ex(2): Find the sum of the given GP series till seven terms? $$ 2 - 4 + 8 - 16.... $$
Solution: From the given series
the first term of the series (a) = 2
Common difference (d) = -2
and n = 7, then $$ S_n = \frac{a \ (d^n - 1)}{d - 1} $$ $$ S_7 = \frac{2 \ ((-2)^7 - 1)}{(-2) - 1} $$ $$ S_7 = \frac{2 \ (- 128 - 1}{- 3} $$ $$ S_7 = \frac{2 \times 129}{3} $$ $$ \frac{258}{3} = 86 $$ Hence the sum of 7 terms of the given GP series is 86.