Question

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

Answer 1

$a=sqrt3$

$r=\frac{3}{\sqrt{3}}=\sqrt{3}$

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

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

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

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

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

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

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

$\therefore n=6$.