(i) According to the given question
G.P: \[1\text{ }-\text{ }1/2\text{ }+\text{ }1/4\text{ }-\text{ }1/8\text{ }+\text{ }\ldots \ldots ..\text{ }to\text{ }9\text{ }terms\]
Here,
\[a\text{ }=\text{ }1\]
And
\[r\text{ }=\text{ }\left( -1/2 \right)/\text{ }1\text{ }=\text{ }-1/2\text{ }\left( \left| \text{ }r\text{ } \right|\text{ }<\text{ }1 \right)\]
Number of terms, \[n\text{ }=\text{ }9\]
Thus,
\[{{S}_{n}}~=\text{ }a(1\text{ }-\text{ }{{r}^{n~}})/\text{ }1\text{ }\text{ }r\]
\[\Rightarrow {{S}_{9}}~=\text{ }\left( 1 \right)(1\text{ }-\text{ }{{\left( -1/2 \right)}^{9~}})/\text{ }\left( 1\text{ }-\text{ }\left( -1/2 \right) \right)\]
\[=\text{ }(1\text{ }+\text{ }{{\left( 1/2 \right)}^{9}})/\text{ }\left( 3/2 \right)\]
\[=\text{ }2/3\text{ }x\text{ }\left( \text{ }1\text{ }+\text{ }1/512\text{ } \right)\]
\[=\text{ }2/3\text{ }x\text{ }\left( 513/512 \right)\]
\[=\text{ }171/\text{ }256\]
(ii) According to the given question
G.P: \[1\text{ }\text{ }1/3\text{ }+\text{ }1/{{3}^{2}}~\text{ }1/{{3}^{3}}~+\text{ }\ldots \ldots \ldots \text{ }to\text{ }n\text{ }terms\]
Here,
\[a\text{ }=\text{ }1\]
And
\[r\text{ }=\text{ }\left( -1/3 \right)/\text{ }1\text{ }=\text{ }-1/3\text{ }\left( \left| \text{ }r\text{ } \right|\text{ }<\text{ }1 \right)\]
Number of terms is \[n\]
Thus,
\[{{S}_{n}}~=\text{ }a(1\text{ }-\text{ }{{r}^{n~}})/\text{ }1\text{ }-\text{ }r\]