Question

The sequence $a_1,a_2,a_3,\dots ,a_{98}$ satisfies the relation $a_{n+1}$ $=a_n+1$ for $n=1,2,3,\dots ,97$ and has the sum equal to 4949. Then the value of $\sum _{k=1}^{49}{a}_{2k}$ 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:
$a_1+\left(a_1+1\right)+\left(a_1+1+1\right)+\dots +\left(a_1+97\right)$
Sum of this series is given $4949$ .
$4949=\frac{98}{2}\left(a_1+a_1+97\right)$
$\Rightarrow 2a_1=101-97$
$\Rightarrow a_1=2$
So, $a_2=3,a_4=5,\dots$
$\therefore a_2+a_4+\dots +a_{98}=3+5+7+\dots +99$
$=\frac{49}{2}(3+99)=2499$