If `a_1, a_2, a_3, ,a_(2n+1)` are in A.P., then `(a_(2n+1)-a_1)/(a_(2n+1)+a_1)+(a_(2n)-a_2)/(a_(2n)+a_2)++(a_(n+2)-a_n)/(a_(n+2)+a_n)` is equal to a. `(n(n+1))/2xx(a_2-a_1)/(a_(n+1))` b. `(n(n+1))/2` c. `(n+1)(a_2-a_1)` d. none of these

Options

A. `(n(n+1))/(2)*(a_(2)-a_(1))/(a_(n+1))`
B. `(n(n+1))/(2)`
C. `(n+1)(a_(2)-a_(1))`
D. none of these

Answer

A

Explanation

In an A.P. the sum of the terms equidistant from the beginning and the end is always same is equal to the sum of first and last term. Therefore,
`a_(1)+a_(2n+1)=a_(2)+a_(2n)= . . .=a_(n)+a_(n+2)`
`:." "(a_(2n+1)-a_(1))/(a_(2n+1)+a_(1))=(a_(2n)-a_(2))/(a_(2n)+a_(2))+ . . .+(a_(n+2)-a_(n))/(a_(n+2)+a_(n))`
`=(2nd+(2n-2)d+ . . .+2d)/(a_(2n+1)+a_(1))`
`=2d(n(n+1))/(2)xx(1)/(a_(2n+1)+a_(1))`
`=2d(n(n+1))/(2)xx(1)/(2a_(n+1))" "[becausea_(1)+a_(2n+1)=a_(n)+a_(n+2)=2a_(n+1)]`
`=d(n(n+1))/(2)xx(1)/(a_(n+1))=(n(n+1))/(2)*(a_(2)-a_(1))/(a_(n+1))" "[becaused=a_(2)-a_(1)]`