# Sum of n terms of a geometric progression:

#### Sum of n terms of a GP series:

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.