Question

If `a_1,a_2,a_3, ,a_n` are an A.P. of non-zero terms, prove that `1/(a_1+a_2)+1/(a_1+a_3)++1/(a_(n-1)+a_n)=(n-1)/(a_1+a_n)dot`

Answer 1

Let a be the common difference of the given AP. Then
`(a_(2) a_(1)) = (a_(3) -a_(2)) = (a_(4) - a_(3)) = …..= ( a_(n) -a_(n-1)) =d`
` 1/ (a_(1)a_(2)) + 1/(a_(2) a_(3)) + 1/(a_(3) a_(4)) +….+ 1/(a_(n-1) a_(n)) `
` = 1/d . { d/(a_(1)a_(2)) + d/(a_(2)a_(3)) + d/ (a_(3) a_(4)) +.....+ d/(a_(n-1) a_(n)) }`
` =1/d .{ ((a_(2) -a_(1))/(a_(1)a_(2)) + ((a_(3) -a_(2))/(a_(2) a_(3)) + ((a_(4) -a_(3)))/(a_(3)a_(4)) +....+((a_(n) -a_(n-1)))/(a_(n-1)a_(n))}`
`[ because(a_(2)-a_(1))=(a_(3)-a_(2))=....=(a_(n)-a_(n-1))=d]`
`=1/d.{(1/a_(1)-1/a_(2))+(1/a_(2)-1/a_(3))+(1/a_(3)-1/a_(4))+....+(1/a_(n-1) -1/a_(n))}`
`= 1/d .[ ({a_(1) + (n-1) d} -a_(1))/(a_(1)a_(n))]`
` 1/d. ((n-1)d)/(a_(1)a_(n))= ((n-1))/(a_(1) a_(n))`
Hence, the result follows.