Question

If $a_1,a_2,a_3,\dots ,a_{2n+1}$ are in $A.P$. then $\frac{a_{2n+1}-a_1}{a_{2n+1}+a_1}+\frac{a_{2n}-a_2}{a_{2n}+a_2}+\cdots +\frac{a_{n+2}-a_n}{a_{n+2}+a_n}$ 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. None of these

Answer 1

Answer: (A)

The general term can be given by
$t_{r+1}=\frac{a_{2n+1-r}-a_{r+1}}{a_{2n+1-r}+a_{r+1}},r=0,1,2,\dots n-1$
$=\frac{a_1+(2n-r)d-\left\{a_1+rd\right\}}{a_1+(2n-r)d+\left\{a_1+rd\right\}}$
$=\frac{(n-r)d}{a_1+nd}$
So, the required sum is
$S_n=\sum _{r=0}^{n-1}{t}_{r+1}$
$=\sum _{r=0}^{n-1}\frac{(n-r)d}{a_1+nd}$
$=\left[\frac{n+(n-1)+(n-2)+\cdots +1}{a_1+nd}\right]d$
$=\frac{n(n+1)d}{2a_{n+1}}$
$=\frac{n(n+1)}{2}\cdot \frac{a_2-a_1}{a_{n+1}}$
$\left[\because d=a_2-a_1\right]$