Question

If $a_1,a_2,a_3,.......,a_{2n+1}$ 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}+.......+ \frac{a_{n+2}-a_n}{a_{n+2}+a_n} $ is equal to

• A. $\frac{n(n+1)}{2},\frac{a_2-a_1}{a_{n+1}} $
• B. $\frac{n(n+1)}{2}$
• C. $(n+1)(a_2-a_1)$
• D. none of these.

Answer 1

Answer: (A)

$\frac{a_{2n}+1 -a_{1}}{a_{2n+1} +a_{1}} =\frac{ a_{1}+\left(2n+1-1\right)d-a_{1}}{a_{1}+\left(2n+1-1\right)d+a_{1}} $
$= \frac{a_{1}+2nd-a_{1}}{a_{1}+2nd+a_{1}} = \frac{nd}{a_{1}+nd} $
$ \frac{a_{n+2}-a_{n}}{a_{n+2} +a_{n}} $
$ = \frac{a_{1} +\left(n+2-1\right)d-\left(a_{1}+\left(n-1\right)d\right)}{a_{1}+\left(n+2-1\right)d +\left(a_{1}+\left(n-1\right)d\right)} $
$ = \frac{\left(n+1\right)d -\left(n-1\right)d}{2a_{1}+\left(n+1\right)d+\left(n-1\right)d} $
$ = \frac{2d}{2a_{1}+2nd} $
$ = \frac{d}{a_{1}+nd} $

$\therefore $ given expression
$ = \frac{d\left(1+2+3+.....n\right)}{a_{1}+nd}$
$ =\frac{ d\cdot n\frac{\left(n+1\right)}{2}}{a_{1}+nd}$