Question

If $a_1, a_2, a_3, \ldots \ldots, a_{2 n+1}$ are in AP then $\frac{a_{2 n+1}-a_1}{a_{2 n+1}+a_1}+\frac{a_{2 n}-a_2}{a_{2 n}+a_2}+\ldots . .+\frac{a_{n+2}-a_n}{a_{n+2}+a_n} \text { is equal to }$

• A. $\frac{n(n+1)}{2} \cdot \frac{a_2-a_1}{a_{n+1}}$
• B. $\frac{n(n+1)}{2}$
• C. $(n+1)\left(a_2-a_1\right)$
• D. $\frac{n(n-1)}{2}\left(\frac{a_2-a_1}{a_{n+1}}\right)$

Answer 1

Answer: (A)

$a_1, a_2, a_3 \ldots \ldots a_{2 n+1}$ are in A.P.
Let common difference is $d$
$a_{2 n+1}=a_1+2 n d \Rightarrow a_{2 n+1}-a_1=2 n d$
Similarly $a_{2 n}-a_2=2(n-1) d$
In denominator, by properties of A.P.
$a_{2 n+1}+a_1=a_{2 n}+a_2=a_{n+2}+a_n$
So all terms in denominator is same
$\text { sum }=\frac{2 d[n+(n-1)+(n-2)+\ldots .1]}{a_{2 n+1}+a_1} $
$\text { sum }=\frac{2 d n(n+1)}{2\left(a_1+a_{2 n+1}\right)}=\frac{n(n+1)\left(a_2-a_1\right)}{a_1+a_{2 n+1}}$
But $a_1+a_{2 n+1}=a_1+a_1+2 n d=2\left(a_1+n d\right)$
$\because a_1+n d=a_{n+1} $
$\therefore a_1+a_{2 n+1}=2 a_{n+1} $
$\therefore \text { sum }=\frac{n(n+1)}{2 a_{n+1}}\left(a_2-a_1\right)$