To solve the problem of finding the value of \((a-b)/(b-c)\) for any three unequal numbers \(a\), \(b\), and \(c\), we can analyze the relationships between these numbers based on the types of progressions they might form.
### Step-by-Step Solution:
1. **Understanding the Problem**: We need to find the expression \((a-b)/(b-c)\). The problem hints at different scenarios where \(a\), \(b\), and \(c\) could be in different types of progressions (AP, GP, HP).
2. **Assuming Arithmetic Progression (AP)**:
- If \(a\), \(b\), and \(c\) are in AP, then we have the property:
\[
2b = a + c
\]
- From this, we can express \(b\) in terms of \(a\) and \(c\):
\[
b = \frac{a+c}{2}
\]
- Now, calculate \(a-b\) and \(b-c\):
\[
a - b = a - \frac{a+c}{2} = \frac{2a - a - c}{2} = \frac{a - c}{2}
\]
\[
b - c = \frac{a+c}{2} - c = \frac{a+c - 2c}{2} = \frac{a - c}{2}
\]
- Now substitute these into the expression:
\[
\frac{a-b}{b-c} = \frac{\frac{a-c}{2}}{\frac{a-c}{2}} = 1
\]
3. **Assuming Geometric Progression (GP)**:
- If \(a\), \(b\), and \(c\) are in GP, we can express them as:
\[
a = q, \quad b = qr, \quad c = qr^2
\]
- Now calculate \(a-b\) and \(b-c\):
\[
a - b = q - qr = q(1 - r)
\]
\[
b - c = qr - qr^2 = qr(1 - r)
\]
- Substitute these into the expression:
\[
\frac{a-b}{b-c} = \frac{q(1-r)}{qr(1-r)} = \frac{1}{r}
\]
- Since \(r = \frac{b}{a}\), we can write:
\[
\frac{a-b}{b-c} = \frac{a}{b}
\]
4. **Assuming Harmonic Progression (HP)**:
- If \(a\), \(b\), and \(c\) are in HP, then we know:
\[
\frac{1}{a}, \frac{1}{b}, \frac{1}{c} \text{ are in AP}
\]
- From the property of AP:
\[
2b = a + c \Rightarrow \frac{2}{b} = \frac{1}{a} + \frac{1}{c}
\]
- Rearranging gives:
\[
\frac{c - b}{bc} = \frac{b - a}{ab}
\]
- Calculate \(a-b\) and \(b-c\):
\[
a - b = a - b, \quad b - c = b - c
\]
- Substitute these into the expression:
\[
\frac{a-b}{b-c} = \frac{ab}{bc} = \frac{a}{c}
\]
### Final Results:
- If \(a\), \(b\), and \(c\) are in AP, then \(\frac{a-b}{b-c} = 1\).
- If \(a\), \(b\), and \(c\) are in GP, then \(\frac{a-b}{b-c} = \frac{a}{b}\).
- If \(a\), \(b\), and \(c\) are in HP, then \(\frac{a-b}{b-c} = \frac{a}{c}\).
