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 terms \( \frac{a}{bc}, \frac{1}{c}, \frac{1}{b} \) are also in AP.

### Step-by-Step Solution:

1. **Understanding AP**:
If \( a, b, c \) are in AP, then the middle term \( b \) is the average of \( a \) and \( c \). This can be expressed mathematically as:
\[ 2b = a + c \]

2. **Rearranging the AP condition**:
From the equation \( 2b = a + c \), we can rearrange it to find a relationship between \( a \), \( b \), and \( c \):
\[ a + c = 2b \]

3. **Dividing by \( bc \)**:
Now, we will divide the entire equation \( a + c = 2b \) by \( bc \):
\[ \frac{a}{bc} + \frac{c}{bc} = \frac{2b}{bc} \]
This simplifies to:
\[ \frac{a}{bc} + \frac{1}{b} = \frac{2}{c} \]

4. **Rearranging the equation**:
We can rearrange the equation to isolate \( \frac{a}{bc} \):
\[ \frac{a}{bc} = \frac{2}{c} - \frac{1}{b} \]

5. **Identifying the terms**:
We have the terms \( \frac{a}{bc}, \frac{1}{c}, \frac{1}{b} \). To show that these are in AP, we need to check if:
\[ 2 \cdot \frac{1}{c} = \frac{a}{bc} + \frac{1}{b} \]

6. **Substituting the expression**:
From the previous step, we know:
\[ \frac{a}{bc} + \frac{1}{b} = \frac{2}{c} \]
Therefore, we can conclude that:
\[ 2 \cdot \frac{1}{c} = \frac{a}{bc} + \frac{1}{b} \]
This confirms that \( \frac{a}{bc}, \frac{1}{c}, \frac{1}{b} \) are indeed in AP.

### Conclusion:
Thus, we have shown that if \( a, b, c \) are in AP, then \( \frac{a}{bc}, \frac{1}{c}, \frac{1}{b} \) are also in AP.