If `log_(x)ax,log_(x)bx" and "log_(x)cx` are in AP, where a, b, c and x, belong to `(1,oo)`, then a, b and c are in

Options

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

Explanation

To solve the problem, we need to show that if \( \log_x a x \), \( \log_x b x \), and \( \log_x c x \) are in Arithmetic Progression (AP), then \( a \), \( b \), and \( c \) are in Geometric Progression (GP).

### Step-by-step Solution:

1. **Understanding the condition of AP**:
Given that \( \log_x a x \), \( \log_x b x \), and \( \log_x c x \) are in AP, we can use the property of AP:
\[ 2 \cdot \log_x b x = \log_x a x + \log_x c x \]

2. **Using the change of base formula**:
We can express the logarithms using the change of base formula:
\[ \log_x a x = \frac{\log a x}{\log x}, \quad \log_x b x = \frac{\log b x}{\log x}, \quad \log_x c x = \frac{\log c x}{\log x} \]

3. **Substituting the logarithmic expressions**:
Substitute these into the AP condition:
\[ 2 \cdot \frac{\log b x}{\log x} = \frac{\log a x}{\log x} + \frac{\log c x}{\log x} \]

4. **Multiplying through by \( \log x \)**:
Multiply both sides by \( \log x \) (since \( x > 1 \), \( \log x > 0 \)):
\[ 2 \log b x = \log a x + \log c x \]

5. **Expanding the logarithms**:
We can expand the logarithms:
\[ 2 (\log b + \log x) = \log a + \log x + \log c + \log x \]
Simplifying gives:
\[ 2 \log b + 2 \log x = \log a + \log c + 2 \log x \]

6. **Canceling \( 2 \log x \)**:
Since \( 2 \log x \) appears on both sides, we can cancel it:
\[ 2 \log b = \log a + \log c \]

7. **Rearranging the equation**:
Rearranging gives:
\[ \log b^2 = \log (a \cdot c) \]

8. **Taking the antilogarithm**:
Taking the antilogarithm of both sides results in:
\[ b^2 = a \cdot c \]

9. **Conclusion**:
The condition \( b^2 = a \cdot c \) indicates that \( a \), \( b \), and \( c \) are in Geometric Progression (GP).

### Final Answer:
Thus, \( a \), \( b \), and \( c \) are in GP.