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. $\frac{n(n+1)}{2}$
• B. $\frac{n(n+1)}{2}\frac{a_2-a_1}{a_{n+1}}$
• C. $(n+1)(a_2-a_1)$
• D. $\frac{n(n+1)}{2}\left(\frac{a_2-a_1}{a_n}\right)$

Answer 1

Answer: (B)

$\frac{n(n+1)}{2}\frac{a_2-a_1}{a_{n+1}}$

Let the common difference be $d=a_2-a_1$ and the first term $a_1=a$
Note that since they are in AP, all denominators of the given series are equal to $a_1+a_{2n+1}=2a+2nd=2(a+nd)=2a_{n+1}$
Also, the numerators are as follows
$2d\left(n\right),2d(n-1),....,2d\left(1\right)$
Hence, on adding all the numerators we get $\frac{n(n+1)}{2}2d$
Hence, the final expression becomes:
$\frac{n(n+1)}{2}\frac{a_2-a_1}{a_{n+1}}$