The sum of 20 terms of the G.P. 10, 20, 40, …… is :

1. 10(2¹⁹ - 1)

2. 10(2²¹ - 1)

3. 10(2²⁰ - 1)

4. none of these

By formula,

Common ratio = $\dfrac{a${n + 1}}{an}

Given,

G.P. = 10, 20, 40, ……

a = 10 and r = $\dfrac{20}{10}$ = 2.

We know that,

If | r | > 1

Sum of n terms of G.P. (S_n) = $\dfrac{a(r^n - 1)}{(r - 1)}$

Substituting values we get :

$\Rightarrow S_{20} = \dfrac{10(2^{20} - 1)}{2 - 1} \\[1em] = \dfrac{10(2^{20} - 1)}{1} \\[1em] = 10(2^{20} - 1).$

Hence, Option 3 is the correct option.