Which term of the GP

Given GP is $\sqrt{3}, 3,3 \sqrt{3} \ldots$

The given GP is of the form, $a, a r, a r^{2}, a r^{3} \ldots .$

Where $r$ is the common ratio.

First term in the given GP, $a_{1}=a=\sqrt{3}$

Second term in GP, $a_{2}=3$

Now, the common ratio, $r=\frac{a_{2}}{a_{1}}$

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

Let us consider 729 as the $n^{\text {th }}$ term of the GP.

Now, $\mathrm{n}^{\text {th }}$ term of GP is, $\mathrm{a}_{\mathrm{n}}=\mathrm{ar}^{\mathrm{n}-1}$

$729=\sqrt{3}(\sqrt{3})^{n-1}$

$\sqrt{3}^{n}=\sqrt{3^{12}}$

$n=12$

So, 729 is the $12^{\text {th }}$ term in GP.