# How many terms of the sequence

Question:

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

Solution:

Here, $a=\sqrt{3}$

Common ratio, $r=\sqrt{3}$

Sum of $n$ terms, $S_{n}=39+3 \sqrt{3}$

$S_{n}=\sqrt{3}\left(\frac{(\sqrt{3})^{n}-1}{\sqrt{3}-1}\right)$

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

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

$\Rightarrow(\sqrt{3})^{n}=1+26$

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

$\Rightarrow(\sqrt{3})^{n}=(\sqrt{3})^{6}$

$\therefore n=6$