Question

Which term of the G.P. :

\[\sqrt{2}, \frac{1}{\sqrt{2}}, \frac{1}{2\sqrt{2}}, \frac{1}{4\sqrt{2}}, . . . \text { is }\frac{1}{512\sqrt{2}}?\]

Answer 1

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

\[\text { and common ratio, }r = \frac{1}{2}\]

\[\text { Let the } n^{th} \text { term be } \frac{1}{512\sqrt{2}} . \]

\[ \therefore a_{n =} \frac{1}{512\sqrt{2}}\]

\[ \Rightarrow a r^{n - 1} = \frac{1}{512\sqrt{2}}\]

\[ \Rightarrow \left( \sqrt{2} \right) \left( \frac{1}{2} \right)^{n - 1} = \frac{1}{512\sqrt{2}}\]

\[ \Rightarrow \left( \frac{1}{2} \right)^{n - 1} = \frac{1}{1024}\]

\[ \Rightarrow \left( \frac{1}{2} \right)^{n - 1} = \left( \frac{1}{2} \right)^{10} \]

\[ \Rightarrow n - 1 = 10 \]

\[ \Rightarrow n = 11\]

\[\text { Thus, the } {11}^{th} \text { term of the given G . P . is } \frac{1}{512\sqrt{2}} .\]