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. A. In AP only
B. B. In GP only
C. C. In both AP and GP
D. D. Neither in AP nor in GP

Answer

To solve the problem step-by-step, we need to analyze the given function and the conditions provided.

### Step 1: Understand the function
The function given is:
\[ f(x) = ax^2 + bx + c \]

### Step 2: Use the condition \( f(1) = f(-1) \)
We need to find the values of \( f(1) \) and \( f(-1) \):
- Calculate \( f(1) \):
\[ f(1) = a(1)^2 + b(1) + c = a + b + c \]
- Calculate \( f(-1) \):
\[ f(-1) = a(-1)^2 + b(-1) + c = a - b + c \]

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

### Step 3: Simplify the equation
By simplifying the equation:
- Cancel \( a \) and \( c \) from both sides:
\[ b + b = 0 \]
\[ 2b = 0 \]
Thus, we find:
\[ b = 0 \]

### Step 4: Use the condition that \( a, b, c \) are in Arithmetic Progression (AP)
Since \( a, b, c \) are in AP, we can express this as:
\[ b = \frac{a + c}{2} \]
Substituting \( b = 0 \):
\[ 0 = \frac{a + c}{2} \]
This implies:
\[ a + c = 0 \]
Thus, we can express \( c \) in terms of \( a \):
\[ c = -a \]

### Step 5: Define the values of \( a, b, c \)
Assuming \( a = k \), then \( c = -k \) and \( b = 0 \). We can choose \( k = 1 \) for simplicity:
- \( a = 1 \)
- \( b = 0 \)
- \( c = -1 \)

### Step 6: Calculate \( f(a), f(b), f(c) \)
Now we need to find:
- \( f(a) = f(1) \):
\[ f(1) = 1(1)^2 + 0(1) - 1 = 1 - 1 = 0 \]
- \( f(b) = f(0) \):
\[ f(0) = 1(0)^2 + 0(0) - 1 = -1 \]
- \( f(c) = f(-1) \):
\[ f(-1) = 1(-1)^2 + 0(-1) - 1 = 1 - 1 = 0 \]

### Step 7: Analyze the values obtained
Now we have:
- \( f(a) = 0 \)
- \( f(b) = -1 \)
- \( f(c) = 0 \)

### Step 8: Determine the type of sequence
The values are \( 0, -1, 0 \).
- To check for Arithmetic Progression (AP), the common difference should be the same:
- From \( f(a) \) to \( f(b) \): \( -1 - 0 = -1 \)
- From \( f(b) \) to \( f(c) \): \( 0 - (-1) = 1 \)

Since the common differences are not the same, they do not form an AP.

- To check for Geometric Progression (GP), the ratio should be the same:
- The ratio from \( f(a) \) to \( f(b) \) is undefined (0 cannot be in the denominator).

Since they do not satisfy the conditions for either AP or GP, we conclude that:
\[ f(a), f(b), f(c) \text{ are neither in AP nor GP.} \]

### Final Answer
Thus, the answer is that \( f(a), f(b), f(c) \) are neither in AP nor GP.

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