If `(1)/(b+c),(1)/(c+a),(1)/(a+b)` are in A.P., then

Options

A. a,b,c are in A.P.
B. `a^(2),b^(2),c^(2)` are in A.P.
C. `(1)/(a),(1)/(b),(1)/( c)` are in A.P.
D. none of these

Answer

To solve the problem, we need to show that if \(\frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b}\) are in arithmetic progression (A.P.), then a certain relationship between \(a\), \(b\), and \(c\) holds.

### Step-by-Step Solution:

1. **Understanding A.P. Condition**:
Since \(\frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b}\) are in A.P., we can use the property of A.P. which states that the middle term is the average of the other two terms. Hence, we can write:
\[ 2 \cdot \frac{1}{c+a} = \frac{1}{b+c} + \frac{1}{a+b} \]

2. **Rearranging the Equation**:
Rearranging the equation gives us:
\[ \frac{2}{c+a} = \frac{1}{b+c} + \frac{1}{a+b} \]

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

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

6. **Expanding Both Sides**:
Expanding both sides:
- Left Side:
\[ 2(ba + 2b^2 + bc + ac) \]
- Right Side:
\[ (c+a)(a + 2b + c) = ca + 2cb + c^2 + a^2 + 2ab + ac \]

7. **Simplifying the Equation**:
After expanding, we can simplify the equation. We will have:
\[ 2ab + 4b^2 + 2bc + 2ac = a^2 + c^2 + 2ab + 2bc + 2ac \]

8. **Cancelling Common Terms**:
Cancelling \(2ab\), \(2bc\), and \(2ac\) from both sides gives:
\[ 4b^2 = a^2 + c^2 \]

9. **Final Result**:
Rearranging this, we find:
\[ 2b^2 = \frac{a^2 + c^2}{2} \]
This indicates that \(a^2, b^2, c^2\) are in A.P.

### Conclusion:
Thus, if \(\frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b}\) are in A.P., then \(a^2, b^2, c^2\) are also in A.P.