Question

If the sum of the first $100$ terms of an arithmetic progression is $-1$ and the sum of the even terms lying in the first $100$ terms is $1,$ then the $100^{\text{th}}$ term of the arithmetic progression is

• A. $\frac{47}{25}$
• B. $\frac{149}{50}$
• C. $\frac{74}{25}$
• D. $-\frac{149}{50}$

Answer 1

Answer: (C)

Given, $S_{100}=-1$
$\Rightarrow \frac{100}{2}\left[2a+99d\right]=-1$
$\Rightarrow 50\left(2a+99d\right)=-1$
Also, $a_2+a_4+a_6+......+a_{100}=1$
$\Rightarrow \frac{50}{2}\left[2\left(a+d\right)+49\left(2d\right)\right]=1$
$\Rightarrow 50\left(a+50d\right)=1$
On solving, we get,
$d=\frac{3}{50}$ and $a=-\frac{149}{50}$
$a_{100}=a+99d=\frac{74}{25}$