Question

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

Answer 1

Given $a_{n+1}=a_n+1 ; n=1,2, \ldots \ldots .97$
$\therefore a_2=a_1+1 $
$a _3= a _1+2 \left( a _3= a _2+1= a _1+1+1= a _1+2\right)$
$\| \text { ly } a_4=a_1+3$
$a _{98}= a _1+97$
hence $a_1, a_2, a_3, \ldots \ldots . . . a_{98}$ form an A.P.
Also given
$\text { Also given } a _1+ a _2+\ldots \ldots . .+ a _{98}=4949 $
$ \frac{98}{2}\left[2 a _1+97 \cdot 1\right]=4949 \text { [using } S _{ n }=\frac{ n }{2}(2 a +\overline{ n -1} d )$
$\Rightarrow 98 a _1+49 \times 97=4949$
$\Rightarrow 98 a _1+4753=4949$
$\therefore 98 a _1=196 \Rightarrow a _1=2$
$\text { now we have to find } $
$ a _2+ a _4+\ldots \ldots+ 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$
$\Rightarrow 49 \times 3+49 \times 48=49(3+48)=49 \times 51=2499$