Let $f\left(x\right)$ be a polynomial function of second degree. If $f\left(1\right)=f(-1)$ and $a,b,c$ are in A.P., then $f\text{'}\left(a\right), f\text{'}\left(b\right), f\text{'}\left(c\right)$ are in 

Let $f\left(x\right)=A{x}^2+Bx+C.$
$f\left(1\right)=f(-1)\Rightarrow A+B+C\Rightarrow A-B+C$
$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2B=0\Rightarrow B=0.\\ \therefore \hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}f\left(x\right)=A{x}^2+C\Rightarrow f^{\mathrm{\prime }}\left(x\right)=2Ax\end{array}$

As $a,b,c$ are in A.P., $2Aa, 2Ab, 2Ac$ are in A.P., i.e., $f\text{'}\left(a\right), f\text{'}\left(b\right), f\text{'}\left(c\right)$ are in A.P.