Question

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

• 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)

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