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

$\sqrt{3},3,3\sqrt{3}$

$S_n=\frac{a(r^n-1)}{r-1}$

$a=\sqrt{3},r=\frac{3}{\sqrt{3}}=\sqrt{3},S_n=39+13\sqrt{3}$

Putting into formula

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

$39+13\sqrt{3}=\frac{({\sqrt{3}}^{n1}-\sqrt{3})}{\sqrt{3}-1}$

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

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

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

$(27\sqrt{3}{)}^1=(\sqrt{3}{)}^{n+1}$

$\left(\sqrt{3}{)}^6\right(\sqrt{3}{)}^1=(\sqrt{3}{)}^{n+1}$

$7=n+1$

$\Rightarrow n=6$