Let `f(x)=ax^(2)+bx+c` such that `f(1)=f(-1)` and a,b,c are in Arithmetic Progression (AP).
Q. `f'(a),f(b),f'(c)` are in

Options

A. AP
B. GP
C. HP
D. Arithmetic-geometric progression

Answer

To solve the problem, we need to analyze the function \( f(x) = ax^2 + bx + c \) under the given conditions: \( f(1) = f(-1) \) and \( a, b, c \) are in Arithmetic Progression (AP).

### Step-by-Step Solution:

1. **Calculate \( f(1) \) and \( f(-1) \)**:
- \( f(1) = a(1)^2 + b(1) + c = a + b + c \)
- \( f(-1) = a(-1)^2 + b(-1) + c = a - b + c \)

2. **Set up the equation from the condition \( f(1) = f(-1) \)**:
- From the condition \( f(1) = f(-1) \), we have:
\[ a + b + c = a - b + c \]

3. **Simplify the equation**:
- Cancel \( a \) and \( c \) from both sides:
\[ b + b = 0 \implies 2b = 0 \]
- Therefore, we find:
\[ b = 0 \]

4. **Check the condition of Arithmetic Progression**:
- Since \( a, b, c \) are in AP, we can express this condition as:
\[ 2b = a + c \]
- Substituting \( b = 0 \) into the equation gives:
\[ 0 = a + c \implies c = -a \]
- This confirms that \( a, 0, -a \) are in AP.

5. **Conclusion**:
- We have determined that \( b = 0 \).

### Final Answer:
The value of \( b \) is \( 0 \).