Find the sum of first 9 terms of the G.P. 1, $-\dfrac{1}{2}$, $\dfrac{1}{4}$, $-\dfrac{1}{8}$, ………

Given,

a = 1

r = $\dfrac{\dfrac{-1}{2}}{1} = -\dfrac{1}{2}$

n = 9

We know that,

The sum of the first n terms of a G.P. is given by:

$S_n = \dfrac{a(1 - r^n)}{1 - r}$ [For r < 1]

Substituting values we get :

$\Rightarrow S_9 = \dfrac{1\Big[1 - \Big(\dfrac{-1}{2}\Big)^9\Big]}{1 - \Big(\dfrac{-1}{2}\Big)} \\[1em] = \dfrac{\Big(1 + \dfrac{1}{512}\Big)}{1 + \dfrac{1}{2}} \\[1em] = \dfrac{\dfrac{512 + 1}{512}}{\dfrac{2 + 1}{2}} \\[1em] = \dfrac{\dfrac{513}{512}}{\dfrac{3}{2}} \\[1em] = \dfrac{513}{512} \times {\dfrac{2}{3}} \\[1em] = \dfrac{171}{256}.$

Hence, S₉ = $\dfrac{171}{256}$.