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

Options

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

Answer

To solve the problem, we need to show that if \( b+c \), \( c+a \), and \( a+b \) are in Harmonic Progression (H.P.), then the fractions \( \frac{b+c}{a} \), \( \frac{c+a}{b} \), and \( \frac{a+b}{c} \) are also in 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.).

Given: \( b+c, c+a, a+b \) are in H.P.
Therefore, we can write:
\[ \frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b} \text{ are in A.P.} \]

2. **Setting Up the A.P. Condition**:
For three numbers to be in A.P., the condition is:
\[ 2 \cdot \frac{1}{c+a} = \frac{1}{b+c} + \frac{1}{a+b} \]
This simplifies to:
\[ 2(c+a) = \frac{(c+a)(a+b) + (c+a)(b+c)}{(b+c)(a+b)} \]

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

4. **Simplifying the Expression**:
We can simplify the right-hand side:
\[ (c+a)(a+b) + (c+a)(b+c) = (c+a)(a+b+b+c) = (c+a)(2b + a + c) \]

5. **Rearranging Terms**:
Rearranging gives us:
\[ 2(b+c)(c+a)(a+b) = (c+a)(2b + a + c) \]

6. **Finding the Required Fractions**:
Now we need to show that \( \frac{b+c}{a}, \frac{c+a}{b}, \frac{a+b}{c} \) are in H.P.
This means we need to show:
\[ \frac{1}{\frac{b+c}{a}}, \frac{1}{\frac{c+a}{b}}, \frac{1}{\frac{a+b}{c}} \text{ are in A.P.} \]

This can be rewritten as:
\[ \frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b} \]

7. **Using the A.P. Condition**:
We need to check if:
\[ 2 \cdot \frac{b}{c+a} = \frac{a}{b+c} + \frac{c}{a+b} \]

8. **Final Verification**:
If we can show that the above condition holds true, then \( \frac{b+c}{a}, \frac{c+a}{b}, \frac{a+b}{c} \) are in H.P.

### Conclusion:
Thus, we conclude that if \( b+c, c+a, a+b \) are in H.P., then \( \frac{b+c}{a}, \frac{c+a}{b}, \frac{a+b}{c} \) are also in H.P.