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Question
Which term of the G.P. :
\[2, 2\sqrt{2}, 4, . . .\text { is }128 ?\]
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Solution
\[\text { Here, first term, } a = 2 \]
\[\text { and common ratio }, r = \sqrt{2}\]
\[\text { Let the } n^{th} \text { term be } 128 . \]
\[ \therefore a_{n =} 128\]
\[ \Rightarrow a r^{n - 1} = 128\]
\[ \Rightarrow \left( 2 \right) \left( \sqrt{2} \right)^{n - 1} = 128\]
\[ \Rightarrow 2 (\sqrt{2} )^{n - 1} = 128\]
\[ \Rightarrow \left( \sqrt{2} \right)^{n - 1} = 64\]
\[ \Rightarrow \left( \sqrt{2} \right)^{n - 1} = \left( \sqrt{2} \right)^{12} \]
\[ \Rightarrow n - 1 = 12 \]
\[ \Rightarrow n = 13\]
\[\text { Thus, the } {13}^{th} \text { term of the given G . P . is } 128 .\]
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