If a, b, c are in HP, then what is \frac{1}{b-a}+\frac{1}{b-c} equal to ? 1. \frac{2}{b} 2. \frac{1}{a}+\frac{1}{c} 3. \frac{1}{2}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) Select the correct answer using the code given below :

Explanation

Since a, b, c are in HP, \frac{1}{a}, \frac{1}{b}, \frac{1}{c} are in AP. Thus \frac{2}{b} = \frac{1}{a} + \frac{1}{c} (Statement 2 is true). The given expression is \frac{1}{b-a} + \frac{1}{b-c}. Substituting b = \frac{2ac}{a+c}, it simplifies to \frac{a+c}{a(c-a)} + \frac{a+c}{c(a-c)} = \frac{a+c}{c-a}(\frac{1}{a}-\frac{1}{c}) = \frac{a+c}{ac} = \frac{1}{a} + \frac{1}{c}. Since this equals \frac{2}{b}, Statement 1 is also true. (Note: The options provided in the original exam are flawed as they do not include '1 and 2 only', but Statement 1 and 2 represent the exact same value).

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