Question

Which term of GP $2,1, \frac{1}{2}, \frac{1}{4}, \ldots$ is $\frac{1}{-128}$?

(A) 8th

(B) 6 th

(C) 7 th

(D) 9 th

Answer 1

Here, $a=2, r=\frac{1}{2}, T_n=\frac{1}{128}$

$\therefore \quad T_n=a r^{n-1} \Rightarrow \frac{1}{128}=2\left(\frac{1}{2}\right)^{n-1}$

$\Rightarrow\left(\frac{1}{2}\right)^{n-1}=\left(\frac{1}{2}\right)^8$

$\Rightarrow \quad n-1=8 $

$\Rightarrow \quad n=9 $

So, The correct option will be (D).