(i) According to the given question

G.P: \[1\text{ }+\text{ }3\text{ }+\text{ }9\text{ }+\text{ }27\text{ }+\text{ }\ldots \ldots \ldots .\text{ }to\text{ }12\text{ }terms\]

Here,

\[a\text{ }=\text{ }1\text{ }and\text{ }r\text{ }=\text{ }3/1\text{ }=\text{ }3\text{ }\left( r\text{ }>\text{ }1 \right)\]

Number of terms, \[~n\text{ }=\text{ }12\]

Thus,

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

\[\Rightarrow {{S}_{12}}~=\text{ }\left( 1 \right)({{\left( 3 \right)}^{12~}}-\text{ }1)/\text{ }3\text{ }\text{ }1\]

\[=\text{ }({{3}^{12}}-\text{ }1)/\text{ }2\]

\[=\text{ }\left( 531441\text{ }-\text{ }1 \right)/\text{ }2\]

\[=\text{ }531440/2\]

\[=\text{ }265720\]

(ii) According to the given question

G.P: \[0.3\text{ }+\text{ }0.03\text{ }+\text{ }0.003\text{ }+\text{ }0.0003\text{ }+\ldots ..~to\text{ }8\text{ }terms\]

Here,

\[a\text{ }=\text{ }0.3\]

And

\[r\text{ }=\text{ }0.03/0.3\text{ }=\text{ }0.1\text{ }\left( r\text{ }<\text{ }1 \right)\]

Number of terms, \[n\text{ }=\text{ }8\]

Thus,

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

\[\Rightarrow {{S}_{8}}~=\text{ }\left( 0.3 \right)(1\text{ }-\text{ }{{0.1}^{8~}})/\text{ }\left( 1\text{ }\text{ }0.1 \right)\]

\[=\text{ }0.3(1\text{ }-\text{ }{{0.1}^{8}})/\text{ }0.9\]

\[=\text{ }(1\text{ }-\text{ }{{0.1}^{8}})/\text{ }3\]

\[=\text{ }1/3(1\text{ }-\text{ }{{\left( 1/10 \right)}^{8}})\]