Find the sum of first 6 terms of the G.P. 1, $-\dfrac{1}{3}, \dfrac{1}{3^{2}}, -\dfrac{1}{3^{3}}$, …….

Given,

a = 1

r = $\dfrac{\dfrac{-1}{3}}{1} = -\dfrac{1}{3}$

n = 6

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 :

$\Rightarrow S_6 = \dfrac{1\Big[1 - \Big(\dfrac{-1}{3}\Big)^6\Big]}{1 - \Big(\dfrac{-1}{3}\Big) } \\[1em] = \dfrac{\Big(1 - \dfrac{1}{3^6}\Big)}{1 + \dfrac{1}{3}} \\[1em] = \dfrac{\Big(1 - \dfrac{1}{729}\Big)}{\dfrac{3 + 1}{3}} \\[1em] = \dfrac{\Big(\dfrac{729 - 1}{729}\Big)}{\dfrac{4}{3}} \\[1em] = \dfrac{\dfrac{728}{729}}{\dfrac{4}{3}} \\[1em] = \dfrac{728}{729}\times \dfrac{3}{4} \\[1em] = \dfrac{182}{243}.$

Hence, S₆ = $\dfrac{182}{243}$.