Correct option is C) \(\frac{-ca}2\)

a, b, c are in A.P.

\(\therefore b = \frac{a + c}2\)   ....(1)

\(\because\) a², b², c² are in H.P.

\(\therefore \frac1{a^2}, \frac 1{b^2}, \frac 1{c^2}\) are in A.P.

\(\therefore \frac1{b^2 } - \frac1{a^2} = \frac1{c^2} - \frac1{b^2}\)

⇒ \(\frac{a^2 -b^2}{a^2b^2} = \frac{b^2 - c^2}{b^2c^2}\)

⇒ \(\frac{(a -b)(a +b)}{a^2} = \frac{(b - c)(b + c)}{c^2}\)

⇒ \(\frac{(a -b)(a +b)}{a^2} = \frac{(a - b)(b + c)}{c^2}\)     \(\begin{pmatrix}\because \text{a, b,c are in A.P.}\\\therefore b -a = c-b\\⇒a -b = b-c\end{pmatrix}\)

⇒ \(\frac{a +b}{a^2} = \frac{b +c}{c^2}\)

⇒ \(a^2 (b + c) = c^2(a + b)\)

⇒ \(a^2b + a^2c = c^2 a + c^2b\)

⇒ \(a^2b - c^2b + a^2c - c^2a = 0\)

⇒ \(b(a^2 - c^2) + ac(a - c) = 0\)

⇒ \(b(a -c)(a + c) + ac(a -c) = 0\)

⇒ \((a -c) (b(a +c) + ac) = 0\)

either \(a -c = 0\, or\, b(a + c) + ac =0\)

If \(a- c = 0\) ⇒ \(a = c\)

which is possible when a = b = c but a, b, c are different

\(\therefore b(a +c) + ac = 0\) 

\(b(2b) + ac = 0 \)  (From(i))

⇒ \(b^2 = \frac{-ac}2\)