If `(1)/(b+c) , (1)/(c+a) and (1)/(a+b)` are in AP, then `a^(2), b^(2) and c^(2)` are in

Options

A. geometric progression
B. arithmetic progression
C. harmonic progression
D. None of these

Explanation

To solve the problem, we need to determine the relationship between \( a^2, b^2, \) and \( c^2 \) given that \( \frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b} \) are in Arithmetic Progression (AP).

### Step-by-Step Solution:

1. **Understanding the Condition for AP:**
The terms \( \frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b} \) are in AP if:
\[ 2 \cdot \frac{1}{c+a} = \frac{1}{b+c} + \frac{1}{a+b} \]

2. **Setting Up the Equation:**
We can rewrite the equation:
\[ 2 \cdot \frac{1}{c+a} = \frac{1}{b+c} + \frac{1}{a+b} \]

3. **Finding a Common Denominator:**
The right-hand side can be combined using a common denominator:
\[ \frac{1}{b+c} + \frac{1}{a+b} = \frac{(a+b) + (b+c)}{(b+c)(a+b)} = \frac{a + 2b + c}{(b+c)(a+b)} \]

4. **Equating the Two Sides:**
Now we equate both sides:
\[ 2 \cdot \frac{1}{c+a} = \frac{a + 2b + c}{(b+c)(a+b)} \]

5. **Cross Multiplying:**
Cross-multiplying gives:
\[ 2(b+c)(a+b) = (a + 2b + c)(c+a) \]

6. **Expanding Both Sides:**
Expanding both sides:
\[ 2(ab + ac + b^2 + bc) = ac + a^2 + 2bc + 2ab + c^2 + 2ac \]

7. **Rearranging the Equation:**
Rearranging leads to:
\[ 0 = a^2 + c^2 + 2ac - 2b^2 \]

8. **Factoring:**
This can be factored as:
\[ a^2 + c^2 + 2ac = 2b^2 \]
or
\[ (a+c)^2 = 2b^2 \]

9. **Conclusion:**
This implies that \( a^2, b^2, c^2 \) are in a specific relationship. Since \( (a+c)^2 = 2b^2 \), we can conclude that \( a^2, b^2, c^2 \) are in **Arithmetic Progression (AP)**.

### Final Answer:
Thus, \( a^2, b^2, c^2 \) are in **AP**.