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Question

Mathematics Question on Sequence and series

If H be the Harmonic mean between a and b, then the value of 1H−a+1H−b\frac{1}{H-a}+ \frac{1}{H-b} is

A

1a−1b\frac{1}{a}-\frac{1}{b}

B

a - b

C

a + b

D

1a+1b\frac{1}{a}+\frac{1}{b}

Answer

1a+1b\frac{1}{a}+\frac{1}{b}

Explanation

Solution

H=2aba+bH= \frac{2ab}{a+b} 1H−a+1H−b=12aba+b−a+12aba+b−b \frac{1}{H-a}+\frac{1}{H-b}= \frac{1}{\frac{2ab}{a+b}-a} + \frac{1}{\frac{2ab}{a+b}-b} =a+b2ab−a2−ab+a+b2ab−ab−b2= \frac{a+b}{2ab-a^{2}-ab}+\frac{a+b}{2ab-ab-b^{2}} =a+bab−a2+a+bab−b2= \frac{a+b}{ab-a^{2}}+ \frac{a+b}{ab-b^{2}} =a+b[1a(b−c)+2b(a−b)]= a+b\left[\frac{1}{a\left(b-c\right)} + \frac{2}{b\left(a-b\right)}\right] =(a+b)[a−bab(a−b)]= \left(a+b\right)\left[\frac{a-b}{ab\left(a-b\right)}\right] =a+bab=1a+1b = \frac{a+b}{ab} = \frac{1}{a}+\frac{1}{b}