For any three unequal numbers a,b and c `(a-b)/(b-c)` equals to

Options

A. `b/c ` id a,b,c are in A.P s
B. `a/b` if a,b,c are in G.P
C. `a/c` if a,b,c are in H.P
D. none of these

Answer

To solve the problem of finding the value of \(\frac{a-b}{b-c}\) for three unequal numbers \(a\), \(b\), and \(c\), we will consider the cases where these numbers are in different types of progressions: Arithmetic Progression (AP), Geometric Progression (GP), and Harmonic Progression (HP).

### Step-by-Step Solution:

**Step 1: Consider the case of Arithmetic Progression (AP)**

In an arithmetic progression, the relationship between the terms can be expressed as:
\[ 2b = a + c \]
From this, we can rearrange it to find \(b\):
\[ b = \frac{a + c}{2} \]

Now, we 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, substituting these into \(\frac{a-b}{b-c}\):
\[ \frac{a - b}{b - c} = \frac{\frac{a - c}{2}}{\frac{a - c}{2}} = 1 \]

**Step 2: Consider the case of Geometric Progression (GP)**

In a geometric progression, the relationship can be expressed as:
\[ b^2 = ac \]

Now, we calculate \(a - b\) and \(b - c\) using the relationship:
Assuming \(b = ar\) and \(c = ar^2\) for some \(a\) and \(r\):
\[ a - b = a - ar = a(1 - r) \]
\[ b - c = ar - ar^2 = ar(1 - r) \]

Now substituting into \(\frac{a-b}{b-c}\):
\[ \frac{a - b}{b - c} = \frac{a(1 - r)}{ar(1 - r)} = \frac{1}{r} \]

**Step 3: Consider the case of Harmonic Progression (HP)**

In a harmonic progression, the relationship can be expressed as:
\[ \frac{2}{b} = \frac{1}{a} + \frac{1}{c} \]
This can be rearranged to find \(b\):
\[ \frac{2}{b} = \frac{c + a}{ac} \implies b = \frac{2ac}{a + c} \]

Now, we calculate \(a - b\) and \(b - c\):
\[ a - b = a - \frac{2ac}{a + c} = \frac{a(a + c) - 2ac}{a + c} = \frac{a^2 - ac}{a + c} \]
\[ b - c = \frac{2ac}{a + c} - c = \frac{2ac - c(a + c)}{a + c} = \frac{2ac - ac - c^2}{a + c} = \frac{ac - c^2}{a + c} \]

Now substituting into \(\frac{a-b}{b-c}\):
\[ \frac{a - b}{b - c} = \frac{\frac{a^2 - ac}{a + c}}{\frac{ac - c^2}{a + c}} = \frac{a^2 - ac}{ac - c^2} \]

### Summary of Results:
- For AP: \(\frac{a-b}{b-c} = 1\)
- For GP: \(\frac{a-b}{b-c} = \frac{1}{r}\)
- For HP: \(\frac{a-b}{b-c} = \frac{a^2 - ac}{ac - c^2}\)