If `b^(2), a^(2), c^(2)` are in A.P., then `a + b, b + c, c + a` will be in

Options

A. G.P.
B. A.P.
C. H.P.
D. none of these

Answer

To solve the problem, we need to show that if \( b^2, a^2, c^2 \) are in Arithmetic Progression (A.P.), then \( a + b, b + c, c + a \) will be in Harmonic Progression (H.P.).

### Step-by-Step Solution:

1. **Understanding the Condition**:
Since \( b^2, a^2, c^2 \) are in A.P., it means that the middle term is the average of the other two terms. Therefore, we can write:
\[ 2a^2 = b^2 + c^2 \]

2. **Rearranging the Equation**:
Rearranging the above equation gives us:
\[ a^2 - b^2 = c^2 - a^2 \]
This indicates that the difference between \( a^2 \) and \( b^2 \) is equal to the difference between \( c^2 \) and \( a^2 \).

3. **Factoring the Differences**:
We can factor the differences:
\[ (a - b)(a + b) = (c - a)(c + a) \]

4. **Cross Multiplying**:
From the factored equation, we can rearrange it:
\[ \frac{a - b}{c + a} = \frac{c - a}{a + b} \]

5. **Adding and Subtracting Terms**:
Now, we can manipulate this equation by adding \( c \) to the left side and subtracting \( c \) from the right side:
\[ \frac{a + c - b}{c + a} = \frac{c - a + b}{a + b} \]

6. **Finding the Harmonic Progression**:
If we denote:
\[ x_1 = a + b, \quad x_2 = b + c, \quad x_3 = c + a \]
We need to show that \( \frac{1}{x_1}, \frac{1}{x_2}, \frac{1}{x_3} \) are in A.P.

7. **Using the Condition**:
From the previous steps, we can see that:
\[ \frac{1}{x_1} + \frac{1}{x_3} = \frac{2}{x_2} \]
This means that \( \frac{1}{x_1}, \frac{1}{x_2}, \frac{1}{x_3} \) are in A.P.

8. **Conclusion**:
Since the reciprocals of \( a + b, b + c, c + a \) are in A.P., it follows that \( a + b, b + c, c + a \) must be in H.P.

### Final Answer:
Thus, we conclude that \( a + b, b + c, c + a \) are in Harmonic Progression (H.P.).