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

Options

A. A.P.
B. GP.
C. H.P.
D. Arithmetico-geometric progression

Explanation

To solve the problem step by step, we will start with the given function and the conditions provided.

### Step-by-Step Solution:

1. **Define the function**:
We have the function \( f(x) = ax^2 + bx + c \).

2. **Use the condition \( f(1) = f(-1) \)**:
Calculate \( f(1) \) and \( f(-1) \):
\[ f(1) = a(1)^2 + b(1) + c = a + b + c \]
\[ f(-1) = a(-1)^2 + b(-1) + c = a - b + c \]
Setting these equal gives:
\[ a + b + c = a - b + c \]
Simplifying this, we can cancel \( a \) and \( c \):
\[ b + b = 0 \implies 2b = 0 \implies b = 0 \]

3. **Substituting \( b \) into the function**:
Now, substituting \( b = 0 \) into the function, we have:
\[ f(x) = ax^2 + c \]

4. **Use the condition that \( a, b, c \) are in Arithmetic Progression (AP)**:
Since \( a, b, c \) are in AP, we have:
\[ 2b = a + c \]
Substituting \( b = 0 \):
\[ 0 = a + c \implies c = -a \]

5. **Rewrite the function**:
Now the function becomes:
\[ f(x) = ax^2 - a = a(x^2 - 1) \]

6. **Find the derivative of the function**:
The derivative \( f'(x) \) is:
\[ f'(x) = \frac{d}{dx}(ax^2 - a) = 2ax \]

7. **Evaluate the derivatives at \( a, b, c \)**:
- For \( f'(a) \):
\[ f'(a) = 2a \cdot a = 2a^2 \]
- For \( f'(b) \) (since \( b = 0 \)):
\[ f'(0) = 2a \cdot 0 = 0 \]
- For \( f'(c) \) (since \( c = -a \)):
\[ f'(-a) = 2a \cdot (-a) = -2a^2 \]

8. **Summarize the results**:
We have:
\[ f'(a) = 2a^2, \quad f'(b) = 0, \quad f'(c) = -2a^2 \]

9. **Check if \( f'(a), f'(b), f'(c) \) are in AP**:
The values are \( 2a^2, 0, -2a^2 \).
To check if they are in AP, we need to verify:
\[ 2 \cdot 0 = 2a^2 + (-2a^2) \implies 0 = 0 \]
This confirms that \( f'(a), f'(b), f'(c) \) are in AP.

### Final Answer:
Thus, \( f'(a), f'(b), f'(c) \) are \( 2a^2, 0, -2a^2 \) and they are in Arithmetic Progression.