According to the given question,

\[{{t}_{4}}~=\text{ }1/27\] and \[{{t}_{7~}}=\text{ }1/729\]

We know that,

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

So,

\[{{t}_{4~}}=\text{ }a{{r}^{4\text{ }-\text{ }1}}~=\text{ }a{{r}^{3}}~=\text{ }1/27\text{ }\ldots .\text{ }\left( 1 \right)\]

\[{{t}_{7~}}=\text{ }a{{r}^{7\text{ }-\text{ }1}}~=\text{ }a{{r}^{6}}~=\text{ }1/729\text{ }\ldots .\text{ }\left( 2 \right)\]

\[Dividing\text{ }\left( 2 \right)\text{ }by\text{ }\left( 1 \right)\text{ }we\text{ }get,\]

\[a{{r}^{6}}/\text{ }a{{r}^{3}}~=\text{ }\left( 1/729 \right)/\text{ }\left( 1/27 \right)\]

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

\[r\text{ }=\text{ }1/3\text{ }\left( r\text{ }<\text{ }1 \right)\]

\[In\text{ }\left( 1 \right)\]

\[a\text{ }x\text{ }1/27\text{ }=\text{ }1/27\]

\[a\text{ }=\text{ }1\]

Thus,

\[{{S}_{n}}~=\text{ }a(1\text{ }-\text{ }{{r}^{n}})/\text{ }1\text{ }\text{ }r\]

\[{{S}_{n}}~=\text{ }(1\text{ }-\text{ }{{\left( 1/3 \right)}^{n~}})/\text{ }1\text{ }-\text{ }\left( 1/3 \right)\]

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

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