According to the given question,

\[{{t}_{5}}~=\text{ }81\text{ }and\text{ }{{t}_{2}}~=\text{ }24\]

We know that,

General term is \[{{t}_{n}}~=\text{ }a{{r}^{n\text{ }-\text{ }1}}\]

So,

\[{{t}_{5}}~=\text{ }a{{r}^{5\text{ }-\text{ }1}}~=\text{ }a{{r}^{4}}~=\text{ }81\text{ }\ldots .\text{ }\left( 1 \right)\]

And,

\[{{t}_{2}}~=\text{ }a{{r}^{2\text{ }-\text{ }1}}~=\text{ }a{{r}^{1}}~=\text{ }24\text{ }\ldots .\text{ }\left( 2 \right)\]

On dividing (1) by (2), we get

\[a{{r}^{4}}/\text{ }ar\text{ }=\text{ }81/\text{ }24\]

\[{{r}^{3}}~=\text{ }27/\text{ }8\]

\[r\text{ }=\text{ }3/2\]

Using \[r\text{ }in\text{ }\left( 2 \right),\]we get

\[a\left( 3/2 \right)\text{ }=\text{ }24\]

\[a\text{ }=\text{ }16\]

Hence, the G.P. is \[G.P.\text{ }=\text{ }a,\text{ }ar,\text{ }a{{r}^{2}},\text{ }a{{r}^{3}}\ldots \ldots \]

\[=\text{ }16,\text{ }16\text{ }x\text{ }\left( 3/2 \right),\text{ }16\text{ }x\text{ }{{\left( 3/2 \right)}^{2}},\text{ }16\text{ }x\text{ }{{\left( 3/2 \right)}^{3}}\]

\[=\text{ }16,\text{ }24,\text{ }36,\text{ }54,\text{ }\ldots \ldots \]