If a, b, c are three distinvt positive real numbers which are in H.P., then $\frac{3 a + 2 b}{2 a - b} + \frac{3 c + 2 b}{2 c - b}$ is

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Solution

The correct option is D

we have $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P. Let $\frac{1}{a}$ = p - q, $\frac{1}{b}$ = p and

$\frac{1}{c}$ = p + q, where p,q > 0 and p > q. Now, substitute these

values in $\frac{3 a + 2 b}{2 a - b} + \frac{3 c + 2 b}{2 c - b}$ then it reduces to

$10 + \frac{14 q^{2}}{p^{2} - q^{2}}$ which is obviously greater than

10(as p > q > 0).

Trick: Put a = 1, b = $\frac{1}{2}$ , c = $\frac{1}{3}$ .

The expression has the value $\frac{3 + 1}{2 - \frac{1}{2}} + \frac{1 + 1}{\frac{2}{3} - \frac{1}{2}}$ = $\frac{8}{3} + 12 > 0$

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