If a,b,c are distinct positive numbers in H.P. then which of the relations must always hold good ?

Options

A. `b/c = (a-b)/(b-c)`
B. `b^(2) gt ac`
C. `b^(2) lt ac`
D. `a/c = (a-b)/(b-c)`

Answer

To determine which relation must always hold good if \(a\), \(b\), and \(c\) are distinct positive numbers in Harmonic Progression (H.P.), we can follow these steps:

### Step-by-Step Solution:

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

2. **Using the Definition of A.P.**:
For three numbers \(x\), \(y\), and \(z\) to be in A.P., the relation \(2y = x + z\) must hold. Therefore, we can write:
\[ 2 \cdot \frac{1}{b} = \frac{1}{a} + \frac{1}{c} \]

3. **Rearranging the Equation**:
Multiplying through by \(abc\) (to eliminate the denominators), we get:
\[ 2ac = bc + ab \]

4. **Rearranging Further**:
Rearranging gives us:
\[ 2ac - ab - bc = 0 \]

5. **Factoring the Equation**:
We can factor this equation as follows:
\[ ac - \frac{ab}{2} - \frac{bc}{2} = 0 \]
or equivalently,
\[ c(b - a) = a(c - b) \]

6. **Dividing Both Sides**:
Dividing both sides by \(ab\) (since \(a\), \(b\), and \(c\) are positive and distinct, \(ab \neq 0\)):
\[ \frac{a}{c} = \frac{b - a}{b - c} \]

7. **Conclusion**:
This relation shows a specific relationship between \(a\), \(b\), and \(c\) when they are in H.P. Thus, we can conclude that:
\[ \frac{a}{c} = \frac{b - a}{b - c} \]
This relation must always hold true.