If \(H_1,H_2,\ldots,H_n\) are n harmonic means between a and b;,then \(\frac{{{H}}_1+a}{{{H}}_1-a}+\frac{{{H}}_n+b}{{{H}}_n-b}\)
Step-by-step Solution:
\[ \frac{1}{a} + d = \frac{1}{H_1} \] and \[ \frac{1}{b} - d = \frac{1}{H_n} \] Now, \[ \frac{H_1 + a}{H_1 - a} + \frac{H_n + b}{H_n - b} = \frac{\frac{1}{a} + \frac{1}{H_1}}{\frac{1}{a} - \frac{1}{H_1}} + \frac{\frac{1}{b} + \frac{1}{H_n}}{\frac{1}{b} - \frac{1}{H_n}} \] \[ = \frac{\frac{1}{a} + \frac{1}{a} + d}{-d} + \frac{\frac{1}{b} + \frac{1}{b} - d}{d} = \frac{2\left( \frac{1}{b} - \frac{1}{a} \right) - 2d}{d} = \frac{2(n + 1)d - 2d}{d} = 2n \]