Question

The sequence $a_1,a_2,a_3,\dots \dots a_{98}$ satisfies the relation $a_{n+1}=a_n+1$ for $n=1,2,3,\dots \dots \dots .97$ and has the sum equal to 4949 . Evaluate $\sum _{k=1}^{49}{a}_{2k}$.

Answer 1

Answer: 2499

Given $a_{n+1}=a_n+1;n=1,2,\dots \dots .97$
$\therefore a_2=a_1+1$
$a_3=a_1+2\left(a_3=a_2+1=a_1+1+1=a_1+2\right)$
$\Vert \text{ ly }{a}_4=a_1+3$
$a_{98}=a_1+97$
hence $a_1,a_2,a_3,\dots \dots ...a_{98}$ form an A.P.
Also given
$\text{ Also given }{a}_1+a_2+\dots \dots ..+a_{98}=4949$
$\frac{98}{2}\left[2a_1+97\cdot 1\right]=4949\text{ [using }{S}_n=\frac{n}{2}(2a+\overline{n-1}d)$
$\Rightarrow 98a_1+49\times 97=4949$
$\Rightarrow 98a_1+4753=4949$
$\therefore 98a_1=196\Rightarrow a_1=2$
$\text{ now we have to find }$
$a_2+a_4+\dots \dots +a_{98}(49\text{ terms })$
$\Rightarrow \frac{49}{2}\left[2\left(a_1+1\right)+49\cdot 2\right]=49\left(a_1+1\right)+49\cdot 48$
$a_1=2$
$\Rightarrow 49\times 3+49\times 48=49(3+48)=49\times 51=2499$