If the mth term of an A.P. is 1/n and the nth term is 1/m, show that the sum of mn terms is 1/2(mn + 1).

Question

If the m^th term of an A.P. is \(1\over n\) and the n^th term is \(1\over m\), show that the sum of mn terms is \(1\over 2\)(mn + 1).

Answer 1

Given that m^th term = \(\frac{1}{n}\) and n^th term = \(\frac{1}{m}\)

To prove: Sum of mn terms of the AP = S_mn = \(\frac{mn + 1}{2}\)

Since m^th term = \(\frac{1}{n}\)

a + (m - 1)d = \(\frac{1}{n}\)

an + mnd - nd = 1 .......(i)

and n^th term = \(\frac{1}{m}\)

a + (n - 1)d = \(\frac{1}{m}\)'

am + mnd - md = 1 ......(ii)