Suppose that m and n are fixed numbers such that the m<sup>th</sup> term of an HP is equal to n and the n<sup>th</sup> term is equal to m, (m ≠ n). Then the (m + n)<sup>th</sup> term is:
The reciprocals of an HP form an AP. Let that AP have first term A and common difference D.
\(A + (m-1)D = \frac{1}{n}\) and \(A + (n-1)D = \frac{1}{m}\).
Subtract: \((m-n)D = \frac{1}{n} - \frac{1}{m} = \frac{m-n}{mn}\), so \(D = \frac{1}{mn}\). Then \(A = \frac{1}{n} - \frac{m-1}{mn} = \frac{1}{mn}\).
\((m+n)\)th term of the AP = \(\frac{1}{mn} + \frac{m+n-1}{mn} = \frac{m+n}{mn}\).
So the \((m+n)\)th term of the HP = \(\frac{mn}{m+n}\).
Answer: (B)
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