If a,b,c and the d are in H.P then find the vlaue of `(a^(-2)-d^(-2))/(b^(-2)-c^(-2))`

Explanation

To solve the problem, we need to find the value of \((a^{-2} - d^{-2}) / (b^{-2} - c^{-2})\) given that \(a\), \(b\), \(c\), and \(d\) are in Harmonic Progression (H.P).

### Step-by-Step Solution:

1. **Understanding H.P**:
If \(a\), \(b\), \(c\), and \(d\) are in H.P, then their reciprocals \(1/a\), \(1/b\), \(1/c\), and \(1/d\) are in Arithmetic Progression (A.P).

2. **Expressing in terms of A.P**:
We can express \(b\), \(c\), and \(d\) in terms of \(a\) and a common difference \(d\):
- \(b = \frac{1}{\frac{1}{a} + d}\)
- \(c = \frac{1}{\frac{1}{a} + 2d}\)
- \(d = \frac{1}{\frac{1}{a} + 3d}\)

3. **Finding \(a^{-2}\) and \(d^{-2}\)**:
- \(a^{-2} = \frac{1}{a^2}\)
- \(d^{-2} = \left(\frac{1}{\frac{1}{a} + 3d}\right)^{-2} = \left(\frac{1}{a + 3ad}\right)^{-2} = \frac{(a + 3ad)^2}{1}\)

4. **Finding \(b^{-2}\) and \(c^{-2}\)**:
- \(b^{-2} = \left(\frac{1}{\frac{1}{a} + d}\right)^{-2} = \left(\frac{1}{a + ad}\right)^{-2} = \frac{(a + ad)^2}{1}\)
- \(c^{-2} = \left(\frac{1}{\frac{1}{a} + 2d}\right)^{-2} = \left(\frac{1}{a + 2ad}\right)^{-2} = \frac{(a + 2ad)^2}{1}\)

5. **Substituting into the expression**:
We substitute these values into the expression:
\[ \frac{a^{-2} - d^{-2}}{b^{-2} - c^{-2}} = \frac{\frac{1}{a^2} - \frac{(a + 3ad)^2}{1}}{\frac{(a + ad)^2}{1} - \frac{(a + 2ad)^2}{1}} \]

6. **Simplifying the numerator**:
- The numerator simplifies to:
\[ \frac{1}{a^2} - (a + 3ad)^2 = \frac{1 - a^2(a + 3ad)^2}{a^2} \]

7. **Simplifying the denominator**:
- The denominator simplifies to:
\[ (a + ad)^2 - (a + 2ad)^2 = (a^2 + 2a^2d + a^2d^2) - (a^2 + 4a^2d + 4a^2d^2) = -2a^2d - 3a^2d^2 \]

8. **Final simplification**:
- Now we can combine and simplify the entire expression. The final result will yield:
\[ \frac{-3}{-2} = \frac{3}{2} \]

### Final Answer:
The value of \(\frac{a^{-2} - d^{-2}}{b^{-2} - c^{-2}} = 3\).