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
Asked by | 16th Sep, 2008, 10:11: PM
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
Answered by | 17th Dec, 2008, 10:15: AM
Kindly Sign up for a personalised experience
- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions
Sign Up
Verify mobile number
Enter the OTP sent to your number
Change