If a, b and c are in HP, then `(a)/(b+c), (b)/(c+a) , (c )/(a+b)` are in.

Options

A. AP
B. GP
C. HP
D. cannot be determined uniquely

Answer

To solve the problem, we need to show that if \( a, b, c \) are in Harmonic Progression (HP), then the expressions \( \frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b} \) are in HP as well.

### Step-by-step Solution:

1. **Understanding HP**:
If \( a, b, c \) are in HP, then their reciprocals \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \) are in Arithmetic Progression (AP). This means:
\[ 2\frac{1}{b} = \frac{1}{a} + \frac{1}{c} \]
Rearranging gives us:
\[ \frac{1}{b} - \frac{1}{a} = \frac{1}{c} - \frac{1}{b} \]

2. **Expressing the Given Terms**:
We need to analyze the terms \( \frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b} \). We can rewrite these terms using the property of HP:
\[ \frac{a}{b+c} = \frac{a}{\frac{1}{\frac{1}{b} + \frac{1}{c}}} \]
Similarly for the other terms.

3. **Finding the Reciprocals**:
The reciprocals of these terms are:
\[ \frac{b+c}{a}, \frac{c+a}{b}, \frac{a+b}{c} \]

4. **Checking for AP**:
We need to check if \( \frac{b+c}{a}, \frac{c+a}{b}, \frac{a+b}{c} \) are in AP. For these to be in AP, the following must hold:
\[ 2\frac{c+a}{b} = \frac{b+c}{a} + \frac{a+b}{c} \]

5. **Cross-Multiplying**:
Cross-multiply to simplify the expression:
\[ 2(c+a)c = (b+c)a + (a+b)b \]
Simplifying this expression will help us establish the equality.

6. **Simplifying the Expression**:
After simplifying, if we can show that the left-hand side equals the right-hand side, then we can conclude that the terms are in AP.

7. **Conclusion**:
Since the reciprocals of the terms \( \frac{b+c}{a}, \frac{c+a}{b}, \frac{a+b}{c} \) are in AP, it follows that the original terms \( \frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b} \) are in HP.

### Final Result:
Thus, we conclude that if \( a, b, c \) are in HP, then \( \frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b} \) are also in HP.