Given,

\[{{t}_{4}}~=\text{ }1/18\text{ }and\text{ }{{t}_{7}}~=\text{ }-1/486\]

General term is

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

So,

\[{{t}_{4}}~=\text{ }a{{r}^{4\text{ }-\text{ }1}}~=\text{ }a{{r}^{3}}~=\text{ }1/18\text{ }\ldots .\text{ }\left( 1 \right)\]

And,

\[{{t}_{7}}~=\text{ }a{{r}^{7\text{ }-\text{ }1}}~=\text{ }a{{r}^{6}}~=\text{ }-1/486\text{ }\ldots .\text{ }\left( 2 \right)\]

Dividing (2) by (1), we get

\[a{{r}^{6}}/\text{ }a{{r}^{3}}~=\text{ }\left( -1/486 \right)/\text{ }\left( 1/18 \right)\]

\[{{r}^{3}}~=\text{ }-1/27\]

\[r\text{ }=\text{ }-1/3\]

Using \[r\text{ }in\text{ }\left( 1 \right),\]we get

\[a{{\left( -1/3 \right)}^{3}}~=\text{ }1/18\]

\[a\text{ }=\text{ }-27/\text{ }18\text{ }=\text{ }-3/2\]

Hence, G.P. is

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

\[=\text{ }-3/2,\text{ }-3/2\left( -1/3 \right),\text{ }-3/2{{\left( -1/3 \right)}^{2}},\]

\[-3/2{{\left( -1/3 \right)}^{3}},\text{ }\ldots \ldots \]

\[=\text{ }-3/2,\text{ }1/2,\text{ }-1/6,\text{ }1/18,\text{ }\ldots ..\]