Solution:

(i) nth term of the G.P. $\sqrt{3}, 1 / \sqrt{3}, 1 / 3 \sqrt{3}, \ldots$
It is known that,
$t_{1}=a=\sqrt{3}, r=t_{2} / t_{1}=(1 / \sqrt{3}) / \sqrt{3}=1 /(\sqrt{3} \times \sqrt{3})=1 / 3$
Using the formula,
$\begin{array}{l}
\mathrm{T}_{n}=\mathrm{ar}^{\mathrm{n}-1} \\
\mathrm{~T}_{\mathrm{n}}=\sqrt{3}(1 / 3)^{\mathrm{n}-1}
\end{array}$

(ii) the $10^{\text {th }}$ term of the G.P. $\sqrt{2}, 1 / \sqrt{2}, 1 / 2 \sqrt{2}, \ldots$
It is known that,
$\mathrm{t}_{1}=\mathrm{a}=\sqrt{2}, \mathrm{r}=\mathrm{t}_{2} / \mathrm{t}_{1}=(1 / \sqrt{2}) / \sqrt{2}=1 /(\sqrt{2} \times \sqrt{2})=1 / 2$
Using the formula,
$\begin{array}{l}
T_{n}=a r^{n-1} \\
T_{10}=\sqrt{2}(1 / 2)^{10-1} \\
=\sqrt{2}(1 / 2)^{9} \\
=1 / \sqrt{2}(1 / 2)^{8}
\end{array}$