If $a, b, c,$ are in A.P and $a^{2}, b^{2}, c^{2}$ are in H.P, then

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

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

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

$a = b = c$

answer is A.

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Detailed Solution

We have $2 b = a = c$

$\begin{array}{l} and b^{2} = \frac{2 a^{2} c^{2}}{a^{2} + c^{2}} \\ {(\frac{a + c}{2})}^{2} = \frac{2 a^{2} c^{2}}{a^{2} + c^{2}} \\ \Rightarrow {(a + c)}^{2} (a^{2} + c^{2}) = 8 a^{2} c^{2} \\ \Rightarrow {(a^{2} + c^{2})}^{2} + 2 a c (a^{2} + c^{2}) - 8 a^{2} c^{2} = 0 \\ \Rightarrow (a^{2} + c^{2} + 4 a c) (a^{2} + c^{2} - 2 a c) = 0 \\ [{(a + c)}^{2} + 2 a c] {[a - c]}^{2} = 0 \\ a - c = 0 \Rightarrow a = c \\ Then a = b = c \end{array}$