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
$=\frac{d\left(1+2+3+.....n\right)}{a_1+nd}$
$=\frac{d\cdot n\frac{\left(n+1\right)}{2}}{a_1+nd}$