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Question

If a1,a2.....an be in AP of non zero terms then prove that
1/(a1a2 ) + 1/(a2a3)+........+1/(an-1an) = (n-1) /(a1an)

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Solution

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


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