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

Options

A. `(n(n+1))/(2)xx(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

The general term can be given by
`t_(r+1)=(a_(2n+1-r)-a_(r+1))/(a_(2n+1-r)+a_(r+1))`,r=0,1,2,..,n-1
`=(a_(1)+(2n-r)d-{a_(1)+rd})/(a_(1)+(2n-r)d+{a_(1)+rd})`
`=((n-r)d)/(a_(1)+nd)`
Therefore, the required sum is
`S_(n)=sum_(r=0)^(n-1)t_(r+1)`
`=sum_(r=0)^(n-1)((n-r)d)/(a_(1)+nd)`
`=[(n+(n-1)+(n-2)+....+1)/(a_(1)+nd)]`
`=(n(n+1)d)/(2a_(n+1))`
`=(n(n+1))/2(a_(2)-1a_(1))/(a_(n+1))` [`becaused=a_(2)-a_(1)`]