If a, b, c be in A.P and `a^(2), b^(2), c^(2)` in H.P., then

Options

A. `-(a)/(2)`, b, c are in G.P.
B. a = b = c
C. a + b + c = 0
D. none

Answer

To solve the problem, we need to analyze the conditions given: \( a, b, c \) are in Arithmetic Progression (A.P.) and \( a^2, b^2, c^2 \) are in Harmonic Progression (H.P.).

### Step-by-Step Solution:

1. **Understanding A.P. Condition**:
Since \( a, b, c \) are in A.P., we can express this as:
\[ b - a = c - b \]
This simplifies to:
\[ 2b = a + c \quad \text{(Equation 1)} \]

2. **Understanding H.P. Condition**:
The numbers \( a^2, b^2, c^2 \) are in H.P. if the reciprocals \( \frac{1}{a^2}, \frac{1}{b^2}, \frac{1}{c^2} \) are in A.P. This means:
\[ 2 \cdot \frac{1}{b^2} = \frac{1}{a^2} + \frac{1}{c^2} \]
Rearranging gives us:
\[ \frac{2}{b^2} = \frac{1}{a^2} + \frac{1}{c^2} \]
Multiplying through by \( a^2b^2c^2 \) gives:
\[ 2a^2c^2 = b^2(c^2 + a^2) \quad \text{(Equation 2)} \]

3. **Substituting from Equation 1**:
From Equation 1, we can express \( c \) in terms of \( a \) and \( b \):
\[ c = 2b - a \]
Now substitute \( c \) into Equation 2:
\[ 2a^2(2b - a)^2 = b^2((2b - a)^2 + a^2) \]

4. **Expanding and Simplifying**:
Expanding both sides:
- Left-hand side:
\[ 2a^2(4b^2 - 4ab + a^2) = 8a^2b^2 - 8a^3b + 2a^4 \]
- Right-hand side:
\[ b^2((4b^2 - 4ab + a^2) + a^2) = b^2(4b^2 - 4ab + 2a^2) = 4b^4 - 4ab^3 + 2a^2b^2 \]

5. **Setting the Equation**:
Equating both sides:
\[ 8a^2b^2 - 8a^3b + 2a^4 = 4b^4 - 4ab^3 + 2a^2b^2 \]
Rearranging gives:
\[ 6a^2b^2 - 8a^3b + 2a^4 - 4b^4 + 4ab^3 = 0 \]

6. **Factoring**:
This equation can be factored or analyzed for roots. One possible solution is to check if \( a = b = c \) holds, which leads to:
\[ a = b = c \]
Thus, \( a, b, c \) can be equal.

7. **Conclusion**:
The values of \( a, b, c \) can be expressed in terms of each other, leading to the conclusion that they can be equal or follow specific ratios based on the conditions of A.P. and H.P.