Question

Which term of the G.P. :

\[2, 2\sqrt{2}, 4, . . .\text {  is }128 ?\]

Answer 1

\[\text { Here, first term, } a = 2 \]

\[\text { and common ratio }, r = \sqrt{2}\]

\[\text { Let the } n^{th} \text { term be } 128 . \]

\[ \therefore a_{n =} 128\]

\[ \Rightarrow a r^{n - 1} = 128\]

\[ \Rightarrow \left( 2 \right) \left( \sqrt{2} \right)^{n - 1} = 128\]

\[ \Rightarrow 2 (\sqrt{2} )^{n - 1} = 128\]

\[ \Rightarrow \left( \sqrt{2} \right)^{n - 1} = 64\]

\[ \Rightarrow \left( \sqrt{2} \right)^{n - 1} = \left( \sqrt{2} \right)^{12} \]

\[ \Rightarrow n - 1 = 12 \]

\[ \Rightarrow n = 13\]

\[\text { Thus, the } {13}^{th} \text { term of the given G . P . is } 128 .\]