Find the sum of first 6 terms of the G.P. 0.1, 0.01, 0.001, …..

Given,

a = 0.1

r = $\dfrac{0.01}{0.1}$ = 0.1

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 :

$\Rightarrow S_6 = \dfrac{0.1[1 - (0.1)^6]}{1 - 0.1} \\[1em] = \dfrac{0.1(1 - 0.000001)}{0.9} \\[1em] = \dfrac{0.1(0.999999)}{0.9} \\[1em] = \dfrac{0.1(0.999999)}{0.9} \\[1em] = \dfrac{1}{9} \times 0.999999 \\[1em] = 0.111111.$

Hence, S₆ = 0.111111.