If `b+c+,c+a,a+b` are in H.P. then which of the following hold(s) good ?

Options

A. `(b+c)/(a),(c+a)/(b), (a+b)/(c )` are in A.P.
B. `(b+c)/(a), (c+a)/(b) , (a+b)/(c )` are in H.P.
C. `a^(2) , b^(2), c^(2)` are in A.P.
D. `a^(2), b^(2), c^(2)` are in H.P.

Answer

To solve the problem, we need to analyze the condition that \( b+c, c+a, a+b \) are in Harmonic Progression (H.P.).

### Step-by-Step Solution:

1. **Understanding H.P.**:
- If three numbers \( x, y, z \) are in H.P., then their reciprocals \( \frac{1}{x}, \frac{1}{y}, \frac{1}{z} \) are in Arithmetic Progression (A.P.).
- Therefore, if \( b+c, c+a, a+b \) are in H.P., it implies that \( \frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b} \) are in A.P.

2. **Setting up the A.P. condition**:
- For three numbers \( x, y, z \) to be in A.P., the condition is:
\[ 2y = x + z \]
- Applying this to our terms, we have:
\[ 2 \cdot \frac{1}{c+a} = \frac{1}{b+c} + \frac{1}{a+b} \]

3. **Cross-multiplying**:
- Cross-multiply to eliminate the fractions:
\[ 2(b+c)(a+b) = (b+c)(c+a) + (c+a)(a+b) \]

4. **Expanding both sides**:
- The left side becomes:
\[ 2(ab + b^2 + ac + bc) \]
- The right side expands to:
\[ (bc + ac + c^2 + ab) + (ca + ab + a^2 + bc) \]
- Simplifying the right side gives:
\[ 2ab + 2bc + ac + a^2 + c^2 \]

5. **Setting the equation**:
- Now we equate both sides:
\[ 2(ab + b^2 + ac + bc) = 2ab + 2bc + ac + a^2 + c^2 \]

6. **Rearranging terms**:
- Rearranging gives:
\[ 2b^2 + 2ac = a^2 + c^2 \]

7. **Final form**:
- This can be rearranged to:
\[ 2b^2 = a^2 + c^2 - 2ac \]
- Recognizing that \( a^2 + c^2 - 2ac = (a - c)^2 \), we have:
\[ 2b^2 = (a - c)^2 \]

8. **Conclusion**:
- This implies that \( a^2, b^2, c^2 \) are in A.P. since \( a - c \) is a constant multiple of \( b \).

### Final Result:
Thus, the conclusion is that \( a^2, b^2, c^2 \) are in A.P.