Solution:
(i) $\sqrt{3}, 3,3 \sqrt{3}, \ldots$ is $729 ?$
Using the formula,
$\begin{array}{l}
\mathrm{T}_{\mathrm{n}}=\mathrm{ar}^{\mathrm{n}-1} \\
\mathrm{a}=\sqrt{3} \\
\mathrm{r}=\mathrm{t}_{2} / \mathrm{t}_{1}=(3 / \sqrt{3}) \\
=\sqrt{3} \\
\mathrm{~T}_{\mathrm{n}}=729 \\
\mathrm{n}=? \\
\mathrm{~T}_{\mathrm{n}}=\mathrm{ar}^{\mathrm{n}-1} \\
729=\sqrt{3}(\sqrt{3})^{\mathrm{n}-1} \\
729=(\sqrt{3})^{\mathrm{n}} \\
3^{6}=(\sqrt{3})^{\mathrm{n}} \\
(\sqrt{3})^{12}=(\sqrt{3})^{\mathrm{n}} \\
\mathrm{n}=12
\end{array}$
Therefore, $12^{\text {th }}$ term of the G.P is 729
(ii) $1 / 3,1 / 9,1 / 27 \ldots$ is $1 / 19683 ?$
Using the formula.
$\begin{array}{l}
T_{n}=a r^{n-1} \\
a=1 / 3 \\
r=t_{2} / t_{1}=(1 / 9) /(1 / 3) \\
=1 / 9 \times 3 / 1
\end{array}$
$\begin{array}{l}
=1 / 3 \\
T_{n}=1 / 19683 \\
n=? \\
T_{n}=a r^{n-1} \\
1 / 19683=(1 / 3)(1 / 3)^{n-1} \\
1 / 19683=(1 / 3)^{n} \\
(1 / 3)^{9}=(1 / 3)^{n} \\
n=9
\end{array}$
Therefore, $9^{\text {th }}$ term of the G.P is $1 / 19683$