If a,b,c are in AP or GP or HP. Then `(a-b)/(b-c)` is equal to:

Options

A. `(b)/(a) or 1 or (b)/(c)`
B. `(c)/(a) or (c)/(b) or 1`
C. `1 or (a)/(b) or (a)/(c)`
D. `1 or (a)/(b) or (c)/(a)`

Answer

To solve the problem, we need to analyze the expression \((a-b)/(b-c)\) under the conditions that \(a\), \(b\), and \(c\) are in Arithmetic Progression (AP), Geometric Progression (GP), or Harmonic Progression (HP).

### Step 1: Analyze the case when \(a\), \(b\), and \(c\) are in AP.
If \(a\), \(b\), and \(c\) are in AP, then we can express the relationship as:
\[ b - a = c - b \]
Let the common difference be \(d\). Therefore, we can write:
\[ b = a + d \]
\[ c = b + d = a + 2d \]

Now, substituting these into the expression \((a-b)/(b-c)\):
\[ a - b = a - (a + d) = -d \]
\[ b - c = (a + d) - (a + 2d) = -d \]
Thus, we have:
\[ \frac{a-b}{b-c} = \frac{-d}{-d} = 1 \]

### Step 2: Analyze the case when \(a\), \(b\), and \(c\) are in GP.
If \(a\), \(b\), and \(c\) are in GP, we can express them as:
\[ b = ar \]
\[ c = ar^2 \]
where \(r\) is the common ratio.

Now substituting these into the expression \((a-b)/(b-c)\):
\[ a - b = a - ar = a(1 - r) \]
\[ b - c = ar - ar^2 = ar(1 - r) \]
Thus, we have:
\[ \frac{a-b}{b-c} = \frac{a(1 - r)}{ar(1 - r)} = \frac{1}{r} \]

### Step 3: Analyze the case when \(a\), \(b\), and \(c\) are in HP.
If \(a\), \(b\), and \(c\) are in HP, then their reciprocals \(1/a\), \(1/b\), and \(1/c\) are in AP. This implies:
\[ \frac{1}{b} - \frac{1}{a} = \frac{1}{c} - \frac{1}{b} \]
Let \(b\) be expressed in terms of \(a\) and \(c\):
\[ b = \frac{2ac}{a+c} \]
Now substituting this into the expression \((a-b)/(b-c)\):
\[ a - b = a - \frac{2ac}{a+c} = \frac{a(a+c) - 2ac}{a+c} = \frac{a^2 + ac - 2ac}{a+c} = \frac{a^2 - ac}{a+c} \]
\[ b - c = \frac{2ac}{a+c} - c = \frac{2ac - c(a+c)}{a+c} = \frac{2ac - ac - c^2}{a+c} = \frac{ac - c^2}{a+c} \]
Thus, we have:
\[ \frac{a-b}{b-c} = \frac{\frac{a^2 - ac}{a+c}}{\frac{ac - c^2}{a+c}} = \frac{a^2 - ac}{ac - c^2} \]
This simplifies to:
\[ \frac{a}{c} \]

### Conclusion:
In summary, we have:
- If \(a\), \(b\), and \(c\) are in AP, \(\frac{a-b}{b-c} = 1\)
- If \(a\), \(b\), and \(c\) are in GP, \(\frac{a-b}{b-c} = \frac{1}{r}\)
- If \(a\), \(b\), and \(c\) are in HP, \(\frac{a-b}{b-c} = \frac{a}{c}\)

Thus, the expression \((a-b)/(b-c)\) can take on different values depending on the type of progression.