Question

How many terms of the geometric sequence $\sqrt{3},3,3\sqrt{3},........$ must be taken to make the sum $39+13\sqrt{3}$?

Answer 1

Given for the geometric series the first term is $\sqrt{3}$ and the common ratio is also $\sqrt{3}$.

Let us take $n$ terms to have the given sum.

Then by the problem,

$\sqrt{3}\frac{(\sqrt{3}{)}^n-1}{\sqrt{3}-1}=39+13\sqrt{3}$

or, $\sqrt{3}\left\{\right(\sqrt{3}{)}^n-1\}=26\sqrt{3}$

or, $(\sqrt{3}{)}^n=27$

or, $n=6$.