If a,b,c are in H.P. , then

Options

A. `(a)/(b + c - a) , (b)/(c + a - b), (c)/(a + b - c)` are in H.P.
B. `(2)/(b) = (1)/(b-a) + (1)/(b-c)`
C. `a - (b)/(2),(b)/(2) , c (b)/(2) ` are in G.P.
D. `(a)/(b+c),(b)/(c+a),(c)/(a +b)` are in H.P.

Answer

To solve the problem, we need to determine the correct statement given that \( a, b, c \) are in Harmonic Progression (H.P.).

### Step-by-Step Solution:

1. **Understanding H.P.**:
- If \( a, b, c \) are in H.P., then their reciprocals \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \) are in Arithmetic Progression (A.P.).
- This means that:
\[ 2b = a + c \]

2. **Analyzing the Options**:
- We will check each option to see if it holds true under the condition \( 2b = a + c \).

3. **Option A**:
- Check if \( \frac{2}{b} = \frac{1}{a} + \frac{1}{c} \).
- The left side is \( \frac{2}{b} \).
- The right side can be expressed as:
\[ \frac{1}{a} + \frac{1}{c} = \frac{c + a}{ac} \]
- Substituting \( a + c = 2b \):
\[ \frac{2b}{ac} \]
- Thus, we need to check if:
\[ \frac{2}{b} = \frac{2b}{ac} \]
- Cross-multiplying gives:
\[ 2ac = 2b^2 \]
- This does not hold true in general, so **Option A is incorrect**.

4. **Option B**:
- Check if \( \frac{2}{b} = \frac{1}{b - c} + \frac{1}{b - a} \).
- The right side becomes:
\[ \frac{(b - a) + (b - c)}{(b - c)(b - a)} = \frac{2b - (a + c)}{(b - c)(b - a)} = \frac{2b - 2b}{(b - c)(b - a)} = 0 \]
- Thus, **Option B is incorrect**.

5. **Option C**:
- Check if \( \frac{a - b}{2} = \frac{b}{2} \).
- This simplifies to \( a - b = b \), or \( a = 2b \).
- This does not necessarily hold true for all \( a, b, c \) in H.P., so **Option C is incorrect**.

6. **Option D**:
- Check if \( \frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b} \) are in H.P.
- Taking reciprocals gives:
\[ \frac{b+c}{a}, \frac{c+a}{b}, \frac{a+b}{c} \]
- We need to show that:
\[ 2 \cdot \frac{c+a}{b} = \frac{b+c}{a} + \frac{a+b}{c} \]
- Using \( a + c = 2b \):
\[ 2 \cdot \frac{2b}{b} = \frac{b + 2b}{a} + \frac{2b + b}{c} \]
- This holds true, confirming that **Option D is correct**.

### Conclusion:
The correct option is **D**: \( \frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b} \) are in H.P.