If a,b,c are in AP, then `(a)/(bc),(1)/(c ), (1)/(b)` are in

Options

A. AP
B. GP
C. HP
D. None of these

Answer

To solve the problem, we need to show that if \( a, b, c \) are in Arithmetic Progression (AP), then the fractions \( \frac{a}{bc}, \frac{1}{c}, \frac{1}{b} \) are also in AP.

### Step-by-step Solution:

1. **Understanding AP**:
If \( a, b, c \) are in AP, it means that:
\[ 2b = a + c \]
This can be rearranged to express \( b \) in terms of \( a \) and \( c \):
\[ b = \frac{a + c}{2} \]

2. **Setting up the fractions**:
We need to check if the terms \( \frac{a}{bc}, \frac{1}{c}, \frac{1}{b} \) are in AP. For these terms to be in AP, the following condition must hold:
\[ 2 \cdot \frac{1}{c} = \frac{a}{bc} + \frac{1}{b} \]

3. **Substituting \( b \)**:
Substitute \( b = \frac{a + c}{2} \) into the equation:
\[ 2 \cdot \frac{1}{c} = \frac{a}{b \cdot c} + \frac{1}{b} \]
We can express \( \frac{1}{b} \) as:
\[ \frac{1}{b} = \frac{2}{a + c} \]

4. **Finding \( \frac{a}{bc} \)**:
Now, we need to find \( \frac{a}{bc} \):
\[ bc = b \cdot c = \left(\frac{a + c}{2}\right) \cdot c = \frac{(a + c)c}{2} \]
Therefore:
\[ \frac{a}{bc} = \frac{a \cdot 2}{(a + c)c} = \frac{2a}{(a + c)c} \]

5. **Combining the terms**:
Now we substitute \( \frac{a}{bc} \) and \( \frac{1}{b} \) into the equation:
\[ 2 \cdot \frac{1}{c} = \frac{2a}{(a + c)c} + \frac{2}{a + c} \]

6. **Finding a common denominator**:
The common denominator for the right side is \( (a + c)c \):
\[ 2 \cdot \frac{1}{c} = \frac{2a + 2c}{(a + c)c} \]
Simplifying gives:
\[ \frac{2}{c} = \frac{2(a + c)}{(a + c)c} \]
This shows that both sides are equal, thus confirming that:
\[ 2 \cdot \frac{1}{c} = \frac{a}{bc} + \frac{1}{b} \]

7. **Conclusion**:
Since the condition for \( \frac{a}{bc}, \frac{1}{c}, \frac{1}{b} \) to be in AP is satisfied, we conclude that:
\[ \frac{a}{bc}, \frac{1}{c}, \frac{1}{b} \text{ are in AP.} \]