Solution:
Given that,
$a_{n}=2 / 3^{n}$
Consider $\mathrm{n}=1,2,3,4, \ldots$ since $\mathrm{n}$ is a natural number.
Therefore,
$\begin{array}{l}
a_{1}=2 / 3 \\
a_{2}=2 / 3^{2}=2 / 9 \\
a_{3}=2 / 3^{3}=2 / 27 \\
a_{4}=2 / 3^{4}=2 / 81
\end{array}$
In $\mathrm{GP}$,
$\begin{array}{l}
a_{3} / a_{2}=(2 / 27) /(2 / 9) \\
=2 / 27 \times 9 / 2 \\
=1 / 3 \\
a_{2} / a_{1}=(2 / 9) /(2 / 3) \\
=2 / 9 \times 3 / 2 \\
=1 / 3
\end{array}$
Therefore, common ratio of consecutive term is $1 / 3$. As a result, $n \in \mathrm{N}$ is a G.P.