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(A) Let the quadratic polynomial be $f(x) = Ax^2 + Bx + C$.Given that $a, b, c$ are in arithmetic progression,we have $2b = a + c$.Given $f(1) = f(-1)

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(A) Let the quadratic polynomial be $f(x) = Ax^2 + Bx + C$.Given that $a, b, c$ are in arithmetic progression,we have $2b = a + c$.Given $f(1) = f(-1)$:$A(1)^2 + B(1) + C = A(-1)^2 + B(-1) + C$$A + B + C = A - B + C$$2B = 0 \implies B = 0$.Thus,$f(x) = Ax^2 + C$.The derivative is $f'(x) = 2Ax$.Then $f'(a) = 2Aa$,$f'(b) = 2Ab$,and $f'(c) = 2Ac$.Since $a, b, c$ are in arithmetic progression,$a, b, c$ satisfy $b - a = c - b$.Multiplying by $2A$,we get $2Ab - 2Aa = 2Ac - 2Ab$,which means $f'(b) - f'(a) = f'(c) - f'(b)$.Therefore,$f'(a), f'(b), f'(c)$ are in arithmetic progression.

Let $f(x)$ be a quadratic polynomial. If $f(1) = f(-1)$ and $a, b, c$ are in arithmetic progression,then $f'(a), f'(b), f'(c)$ are in..... progression.

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