2016/033) If $a,b,c$ are in AP and $a^2,b^2,c^2$ are in HP then prove that either $a=b=c$ or $a,b,\frac{-c}{2}$ are in GP

we have $2b = a + c\cdots(1)$
$\frac{2}{b^2} = \frac{1}{a^2} + \frac{1}{c^2}$
or  $2a^2c^2 = b^2c^2 + b^2 a^2 = b^2(a^2+c^2) = b^2((a+c)^2 - 2ac) = b^2(4b^2 - 2ac)$
or $a^2c^2 = 2b^4 - b^2ac$
or $2b^4 - b^2ac - a^2c^2 = (2b^2+ac)(b^2-ac) = 0$
$b^2= ac =>  4b^2 = 4ac = (a+c)^2 => (a-c)^2 = 0 => a = c = b$
or $2b^2+ ac = 0 => a,b ,\frac{-c}{2}$ are in GP