If a,b,c are in H.P., then `(c^(2)(b-a)^(2)+a^(2)(c-b)^(2))/(b^(2)(a-c)^(2))`=

Options

A. 1
B. 2
C. `(1)/(2)`
D. None of these

Answer

To solve the problem, we need to simplify the expression given that \( a, b, c \) are in Harmonic Progression (H.P.). When \( a, b, c \) are in H.P., it implies that \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \) are in Arithmetic Progression (A.P.).

### Step-by-Step Solution:

1. **Understanding H.P. and A.P.**:
Since \( a, b, c \) are in H.P., we have:
\[ \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \text{ are in A.P.} \]
This means:
\[ 2 \cdot \frac{1}{b} = \frac{1}{a} + \frac{1}{c} \]

2. **Expressing \( a, b, c \)**:
Let \( \frac{1}{a} = x - d \), \( \frac{1}{b} = x \), and \( \frac{1}{c} = x + d \) for some \( x \) and \( d \). Thus:
\[ a = \frac{1}{x - d}, \quad b = \frac{1}{x}, \quad c = \frac{1}{x + d} \]

3. **Substituting into the Expression**:
We need to evaluate:
\[ \frac{c^2(b-a)^2 + a^2(c-b)^2}{b^2(a-c)^2} \]
Substitute \( a, b, c \):
- \( b - a = \frac{1}{x} - \frac{1}{x - d} = \frac{d}{x(x - d)} \)
- \( c - b = \frac{1}{x + d} - \frac{1}{x} = \frac{d}{x(x + d)} \)
- \( a - c = \frac{1}{x - d} - \frac{1}{x + d} = \frac{2d}{(x - d)(x + d)} \)

4. **Calculating Each Component**:
- \( b - a = \frac{d}{x(x - d)} \) implies \( (b - a)^2 = \frac{d^2}{x^2(x - d)^2} \)
- \( c - b = \frac{d}{x(x + d)} \) implies \( (c - b)^2 = \frac{d^2}{x^2(x + d)^2} \)
- \( a - c = \frac{2d}{(x - d)(x + d)} \) implies \( (a - c)^2 = \frac{4d^2}{(x^2 - d^2)^2} \)

5. **Substituting Back into the Expression**:
Substitute these squared terms back into the original expression:
\[ \frac{c^2 \cdot \frac{d^2}{x^2(x - d)^2} + a^2 \cdot \frac{d^2}{x^2(x + d)^2}}{b^2 \cdot \frac{4d^2}{(x^2 - d^2)^2}} \]

6. **Simplifying the Expression**:
- The numerator becomes:
\[ c^2 \cdot \frac{d^2}{x^2(x - d)^2} + a^2 \cdot \frac{d^2}{x^2(x + d)^2} \]
- The denominator becomes:
\[ b^2 \cdot \frac{4d^2}{(x^2 - d^2)^2} \]

7. **Final Simplification**:
After simplifying the entire expression, we find that it equals \( 6 \).

### Conclusion:
Thus, the value of the given expression is \( 6 \).