The harmonic mean \(H\) of any two numbers a and \(b\) is a number such that the numbers  \(\frac{1}{a},\frac{1}{H},\frac{1}{b}\) are in \(A.P.\) Thus, we have

\(\frac{2}{H} = \frac{1}{a} + \frac{1}{b}\,\,\,\,\,\)or \(\,H = \frac{{2ab}}{{a + b}}\) 

The \(n\) numbers \({H_1},{H_2}, \ldots  \ldots  \ldots .{H_n}\) are said to be \(n\) harmonic means between \(a\) and \(b\) if \(\frac{1}{a},\,\,\frac{1}{{{H_1}}},\,\,\frac{1}{{{H_2}}},\,\,..............,\,\,\frac{1}{{{H_n}}},\,\,\frac{1}{b}\) is an \(A.P.\). Here \(\frac{1}{a}\) is the first term and \(\frac{1}{b}\)is the \({\left( {n + {\rm{ }}2} \right)^{th}}\) term of the \(A.P.\)

If \(d\) is the common difference of this A.P., then we have 

\(\frac{1}{b} = \frac{1}{a} + (n + 2 - 1)d\,\)\( \Rightarrow \,d = \frac{{a - b}}{{(n + 1)ab}}\)

Example 20: Insert 2 Harmonic Means between the numbers \(\frac{1}{3}\) and \(\frac{1}{9}\)

Solution: Suppose the two harmonic means are \({H_1},{H_2}\). Now 3, \(\frac{1}{{{H_1}}},\,\,\,\frac{1}{{{H_2}}}\), 9 are in \(AP\). Whose first term is 3 and the fourth term is 9, hence

\(3 + 3d = 9\) or \(d = 2\)

Hence \(\frac{1}{{{H_1}}}= 5\) and \(\frac{1}{{{H_2}}}= 7\) or \({H_1} = \frac{1}{5}\), \({H_2} \frac{1}{7}\)