Question

If $a_{1}, a_{2}, a_{3}, \ldots, a_{2 n+1}$ are in $A.P$. then $\frac{a_{2 n+1}-a_{1}}{a_{2 n+1}+a_{1}}+\frac{a_{2 n}-a_{2}}{a_{2 n}+a_{2}}+\cdots+\frac{a_{n+2}-a_{n}}{a_{n+2}+a_{n}}$ 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
$t_{r +1}=\frac{a_{2 n+1-r}-a_{r+1}}{a_{2 n+1-r}+a_{r+1}}, r=0,1,2, \ldots n-1$
$=\frac{a_{1}+(2 n-r) d-\left\{a_{1}+r d\right\}}{a_{1}+(2 n-r) d+\left\{a_{1}+r d\right\}}$
$=\frac{(n-r) d}{a_{1}+n d}$
So, the required sum is
$S_{n} =\displaystyle\sum_{r=0}^{n-1} t_{r+1}$
$=\displaystyle\sum_{r=0}^{n-1} \frac{(n-r) d}{a_{1}+n d}$
$=\left[\frac{n+(n-1)+(n-2)+\cdots+1}{a_{1}+n d}\right] d$
$=\frac{n(n+1) d}{2 a_{n+1}}$
$=\frac{n(n+1)}{2} \cdot \frac{a_{2}-a_{1}}{a_{n+1}}$
$\left[\because d=a_{2}-a_{1}\right]$