Question

The sequence $a_{1}, a_{2}, a_{3}, \ldots, a_{98}$ satisfies the relation $a_{n+1}$ $=a_{n}+1$ for $n=1,2,3, \ldots, 97$ and has the sum equal to 4949. Then the value of $\displaystyle\sum_{k=1}^{49} a_{2 k}$ is

• A. 2501
• B. 2499
• C. 2401
• D. 2110

Answer 1

Answer: (B)

We have $ a_{n+1}=a_{n}+1$
$a_{2}=a_{1}+1 $
$a_{3}=a_{2}+1$
Thus, the series is:
$a_{1}+\left(a_{1}+1\right)+\left(a_{1}+1+1\right)+\ldots+\left(a_{1}+97\right)$
Sum of this series is given $4949$ .
$4949=\frac{98}{2}\left(a_{1}+a_{1}+97\right)$
$\Rightarrow 2 a_{1}=101-97$
$\Rightarrow a_{1}=2$
So, $a_{2}=3, a_{4}=5, \ldots$
$ \therefore a_{2}+a_{4}+\ldots+a_{98} =3+5+7+\ldots+99 $
$=\frac{49}{2}(3+99)=2499 $