The seventh term in the sequence 1, -1/2,1/4……………A) 1/64B) -

1.	The seventh term in the sequence 1, -1/2,1/4……………A) 1/64B) -1/8C) 1/16D) -1/32
Answer» 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. \(=a_7=ar^6\) \(=1.(\frac{-1}2)^6=\frac1{2^6}\) \(=\frac1{64}\) Correct option is A) 1/64

The seventh term in the sequence 1, -1/2,1/4……………A) 1/64B) -1/8C) 1/16D) -1/32

Answer»

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. \(=a_7=ar^6\)

\(=1.(\frac{-1}2)^6=\frac1{2^6}\)

\(=\frac1{64}\)

Correct option is A) 1/64