Question

The sum of $20$ terms of the progression $\frac{1}{4},-\frac{1}{2},1,-2,4,\dots \dots$ is

• A. $-\frac{1}{12}\left[\right(2{)}^{20}-1]$
• B. $\frac{1}{12}\left[\right(2{)}^{20}-1]$
• C. $\frac{1}{4}\left[\right(2{)}^{20}-1]$
• D. $-\frac{1}{4}\left[\right(2{)}^{20}-1]$

Answer 1

The correct option is A $-\frac{1}{12}\left[\right(2{)}^{20}-1]$

In the given series, we have
$-\frac{1}{2}÷\left(\frac{1}{4}\right)=-2,1÷(-\frac{1}{2})=-2$
So, the given progression is a G.P. in which $a=\frac{1}{4}$ and $r=-2$.
$\therefore S_{20}=\frac{\left(1/4\right)\left[\right(-2{)}^{20}-1]}{-2-1}=-\frac{1}{12}\left[\right(2{)}^{20}-1]$