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

Options

A. `(a-b)/(b-c)=(a)/(c)`
B. `(b-c)/(c-a)=(b)/(a)`
C. `(c-a)/(a-b)=(c)/(b)`
D. `(a-b)/(b-c)=(c)/(a)`

Answer

To solve the problem, we need to establish the relationship between \( a \), \( b \), and \( c \) when they are in Harmonic Progression (H.P).

### Step-by-Step Solution:

1. **Understanding Harmonic Progression**:
If \( a \), \( b \), and \( c \) are in Harmonic Progression, then the reciprocals \( \frac{1}{a} \), \( \frac{1}{b} \), and \( \frac{1}{c} \) are in Arithmetic Progression (A.P).

2. **Setting up the A.P Condition**:
The condition for three numbers \( x, y, z \) to be in A.P is given by:
\[ 2y = x + z \]
Applying this to our reciprocals, we have:
\[ 2 \cdot \frac{1}{b} = \frac{1}{a} + \frac{1}{c} \]

3. **Rearranging the Equation**:
We can rewrite the equation:
\[ \frac{2}{b} = \frac{1}{a} + \frac{1}{c} \]
To combine the fractions on the right side, we find a common denominator:
\[ \frac{2}{b} = \frac{c + a}{ac} \]

4. **Cross-Multiplying**:
Now, we cross-multiply to eliminate the fractions:
\[ 2ac = b(a + c) \]

5. **Rearranging for Further Analysis**:
We can rearrange this equation to express a relationship among \( a \), \( b \), and \( c \):
\[ \frac{a}{c} = \frac{a - b}{b - c} \]

6. **Conclusion**:
From the derived relationship, we conclude that:
\[ \frac{a}{c} = \frac{a - b}{b - c} \]
Therefore, the correct option is \( a \).