If a,b,c are in H.P. then

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$\frac{a}{b + c}, \frac{b}{c + a}, \frac{c}{a + b}$ are in H.P

$\frac{2}{b} = \frac{1}{b - c} + \frac{1}{b - c}$

$\frac{a}{b + c - a}, \frac{b}{c + a - b}, \frac{c}{a + b - c}$ are in H.P

$a - \frac{b}{2}, \frac{b}{2}, c - \frac{b}{2}$ are in G.P

answer is A, B, C, D.

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Detailed Solution

$\frac{1}{a}, \frac{1}{b}, \frac{1}{c} are in A.P. \Rightarrow \frac{a + b + c}{a}, \frac{a + b + c}{b}, \frac{a + b + c}{c} are in A.P. \Rightarrow \frac{b + c}{a}, \frac{c + a}{b}, \frac{a + b}{c} are in A.P. (Subtracting 1 from each term) \Rightarrow \frac{b + c}{a} - 1, \frac{c + a}{b} - 1, \frac{a + b}{c} - 1 are in A.P. (Subtracting 1 from each term) \Rightarrow \frac{b + c - a}{a}, \frac{c + a - b}{b}, \frac{a + b - c}{c} are in A.P.$

$a, b, c are in H . P \Rightarrow b = \frac{2 ac}{a + c} \frac{1}{b - a} + \frac{1}{b - c} = \frac{2 b - (a + c)}{(b - a) (b - c)} = \frac{2 b - (a + c)}{b^{2} - b (a + c) + a c} = \frac{2 b - 2 a c / b}{b^{2} - b (2 a c) b + a c} = \frac{2}{b} \cdot \frac{b^{2} - a c}{b^{2} - a c} = \frac{2}{b}$

a,b,c are in H.P

$b = \frac{2 a c}{a + c} \Rightarrow a b + b c = 2 a c \Rightarrow 2 a b + 2 b c = 4 a c \Rightarrow 2 a b + 2 b c + b^{2} = 4 a c + b^{2} \Rightarrow b^{2} = 4 a c - 2 a b - 2 b c + b^{2} \Rightarrow b^{2} = 2 a (2 c - b) - b (2 c - b) \Rightarrow b^{2} = (2 c - b) (2 a - b) \Rightarrow {(\frac{b}{2})}^{2} = (a - \frac{b}{2}) (c - \frac{b}{2})$