Correct option is (C) 5
Let \(a_1=2,a_2=\frac{-1}{\sqrt2},a_3=\frac14\)
\(\because\frac{a_2}{a_1}=\frac{\frac{-1}{\sqrt2}}2=\frac{-1}{2\sqrt2},\)
\(\frac{a_3}{a_2}=\frac{\frac{1}{4}}{\frac{-1}{\sqrt2}}=\frac{-\sqrt2}{4}=\frac{-1}{2\sqrt2}\)
\(\because\frac{a_3}{a_2}=\frac{a_2}{a_1}\)
\(\therefore\) \(2,\frac{-1}{\sqrt2},\frac{1}{4},-......\) will form a G.P. with \(r=\frac{-1}{2\sqrt2}\)
Let \(n^{th}\) term of the G.P. be \(\frac1{32}.\)
\(\therefore a_n=\frac1{32}\)
\(\Rightarrow ar^{n-1}=\frac1{32}\)
\(\Rightarrow2(\frac{-1}{2\sqrt2})^{n-1}=\frac1{32}\)
\(\Rightarrow\left(\frac{-1}{2\sqrt2}\right)^{n-1}=\frac1{32\times2}=\frac1{64}\)
\(\Rightarrow\left(\frac{-1}{2\sqrt2}\right)^{n-1}=(\frac18)^2=(\frac{-1}{2\sqrt2})^4\) \(\left(\because\left(\frac{-1}{2\sqrt2}\right)^{2}=\frac18\right)\)
\(\therefore n-1=4\) (By comparing)
\(\Rightarrow n=4+1=5\)
Hence, \(5^{th}\) term of \(2-\frac{1}{\sqrt2}+\frac{1}{4}-.......\) is \(\frac1{32}.\)
