If a1,a2,a3.....are in A.P Prove that ::1/a1.a2 + 1/a2.a3 + 1/a3.a4 +...+ 1/a(n-1).a(n)=(n-1)/a1.an

let first term = a

and common difference = d

then a1 = a, a2 =a+d, a3= a+2d.....

1/a1.a2  + 1/a2.a3  = 1/a(a+d)  + 1/((a+d)(a+2d)  = (a+2d+a)/a(a+d)(a+2d) = 2/a(a+2d) = 2/a1.a3

now 2/a1.a3 + 1/a3.a4  = 2/a(a+2d)  +1/(a+2d)(a+3d)  = (2a+6d+a) /a(a+2d)(a+3d) = 3/a(a+3d) = 3/a1.a4

so if in denominator we have an then in numerator we will be having (n-1).

continuing this the sum of given expression will be = (n-1) /a1.an