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. in A.P. only
B. in GP. Only
C. in both A.P. and G.P.
D. neither in A.P. nor in G.P.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the given function
The function is given as:
\[ f(x) = ax^2 + bx + c \]
We also know that \( f(1) = f(-1) \) and that \( a, b, c \) are in Arithmetic Progression (AP).

### Step 2: Set up the equation from the condition \( f(1) = f(-1) \)
Calculating \( f(1) \):
\[ f(1) = a(1)^2 + b(1) + c = a + b + c \]

Calculating \( f(-1) \):
\[ f(-1) = a(-1)^2 + b(-1) + c = a - b + c \]

Setting these two equal gives us:
\[ a + b + c = a - b + c \]

### Step 3: Simplify the equation
Subtract \( a + c \) from both sides:
\[ b = -b \]

Adding \( b \) to both sides:
\[ 2b = 0 \]
Thus, we find:
\[ b = 0 \]

### Step 4: Use the fact that \( a, b, c \) are in AP
Since \( a, b, c \) are in AP and we found \( b = 0 \), we can express \( c \) in terms of \( a \):
- The condition for \( a, b, c \) to be in AP is that \( 2b = a + c \).
- Substituting \( b = 0 \) gives:
\[ 0 = a + c \]
Thus:
\[ c = -a \]

### Step 5: Write the function with the values of \( b \) and \( c \)
Now, substituting \( b \) and \( c \) back into the function:
\[ f(x) = ax^2 + 0 \cdot x - a = ax^2 - a \]

### Step 6: Find the second derivative of \( f(x) \)
The first derivative \( f'(x) \):
\[ f'(x) = 2ax \]

The second derivative \( f''(x) \):
\[ f''(x) = 2a \]

### Step 7: Evaluate \( f''(a), f''(b), f''(c) \)
Since the second derivative \( f''(x) \) is constant:
- \( f''(a) = 2a \)
- \( f''(b) = 2a \)
- \( f''(c) = 2a \)

### Conclusion
Thus, we conclude:
\[ f''(a) = f''(b) = f''(c) = 2a \]