Correct option is (A) 1/64
Given sequence is \(1 , \frac{-1}{2}, \frac{1}{4}, ........\)
\(\therefore a_1=1,a_2=\frac{-1}2,a_3=\frac14\)
Now, \(\frac{a_2}{a_1}=\frac{\frac{-1}2}1=\frac{-1}2,\)
\(\frac{a_3}{a_2}=\cfrac{\frac14}{\frac{-1}2}\)
\(=\frac14\times-2=\frac{-1}2\)
\(\because\) \(\frac{a_3}{a_2}=\frac{a_2}{a_1}\)
\(\therefore\) \(1 , \frac{-1}{2}, \frac{1}{4}, ........\) will form a G.P. with common difference \(\frac{-1}2.\)
\(\therefore r=\frac{a_2}{a_1}=\frac{-1}2\)
\(\therefore7^{th}\) term of G.P. \(=a_7=ar^6\)
\(=1.(\frac{-1}2)^6=\frac1{2^6}\)
\(=\frac1{64}\)
