Question

Which term of the sequence $2,1,\frac{1}{2},\frac{1}{4},\dots$ is $\frac{1}{128}$

• A. $9$
• B. $8$
• C. $10$
• D. $7$

Answer 1

The correct option is A $9$

Clearly, the given progression is a G.P. with first term $a=2$ and common ratio $r=\frac{1}{2}$.
Let the $n^{\text{th}}$ term be $\frac{1}{128}$. Then,
$a_n=\frac{1}{128}$
$\Rightarrow a{r}^{n-1}=\frac{1}{128}$
$\Rightarrow 2{\left(\frac{1}{2}\right)}^{n-1}=\frac{1}{128}$
$\Rightarrow {\left(\frac{1}{2}\right)}^{n-2}={\left(\frac{1}{2}\right)}^7$
$\Rightarrow n-2=7$
$\Rightarrow n=9$
Thus, 9th term of the given G.P. is $\frac{1}{128}$.