$f\left( x \right) = a{x^2} + bx + c$ $f\left( 1 \right) = f\left( { - 1} \right)$ $\Rightarrow a + b + c = a - b + c$ or $b = 0$ $\therefore$ $f\left( x \right) = a{x^2} + c$ or $f'\left( x \right) = 2ax$ Now $f'\left( a \right);f'\left( b \right);$ and $f'\left( c \right)$ are $2a\left( a \right);2a\left( b \right);2a\left( c \right)$ i.e.$\,2{a^2},\,2ab,\,2ac.$ $\Rightarrow$ If $a,b,c$ are in $A.P.$ then $f'\left( a \right);f'\left( b \right)$ and $f'\left( c \right)$ are also in $A.P.$