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(D) Given that $a, b, c$ are in $H.P.$,their reciprocals $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in $A.P.$Let $\frac{1}{a} = p - q, \frac{1}{b} =

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(D) Given that $a, b, c$ are in $H.P.$,their reciprocals $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in $A.P.$Let $\frac{1}{a} = p - q, \frac{1}{b} = p, \frac{1}{c} = p + q$,where $p, q > 0$ and $p > q$.Then $a = \frac{1}{p-q}, b = \frac{1}{p}, c = \frac{1}{p+q}$.Substituting these into the expression $E = \frac{3a + 2b}{2a - b} + \frac{3c + 2b}{2c - b}$:$E = \frac{\frac{3}{p-q} + \frac{2}{p}}{\frac{2}{p-q} - \frac{1}{p}} + \frac{\frac{3}{p+q} + \frac{2}{p}}{\frac{2}{p+q} - \frac{1}{p}}$$E = \frac{3p + 2(p-q)}{2p - (p-q)} + \frac{3p + 2(p+q)}{2p - (p+q)} = \frac{5p - 2q}{p + q} + \frac{5p + 2q}{p - q}$$E = \frac{(5p - 2q)(p - q) + (5p + 2q)(p + q)}{p^2 - q^2} = \frac{5p^2 - 7pq + 2q^2 + 5p^2 + 7pq + 2q^2}{p^2 - q^2} = \frac{10p^2 + 4q^2}{p^2 - q^2}$$E = \frac{10(p^2 - q^2) + 14q^2}{p^2 - q^2} = 10 + \frac{14q^2}{p^2 - q^2}$.Since $p > q > 0$,$p^2 - q^2 > 0$,so $E > 10$.

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

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