If a,b,c are in GP then `1/(log_(a)x), 1/(log_(b)x), 1/(log_( c)x)` 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 geometric progression (GP), then \( \frac{1}{\log_a x}, \frac{1}{\log_b x}, \frac{1}{\log_c x} \) are in arithmetic progression (AP).

### Step-by-Step Solution:

1. **Understanding the GP Condition**:
Since \( a, b, c \) are in GP, we have the condition:
\[ b^2 = ac \]

2. **Using the Logarithmic Identity**:
We know that:
\[ \log_a x = \frac{\log x}{\log a}, \quad \log_b x = \frac{\log x}{\log b}, \quad \log_c x = \frac{\log x}{\log c} \]
Therefore, we can express:
\[ \frac{1}{\log_a x} = \frac{\log a}{\log x}, \quad \frac{1}{\log_b x} = \frac{\log b}{\log x}, \quad \frac{1}{\log_c x} = \frac{\log c}{\log x} \]

3. **Rewriting the Terms**:
We can rewrite our terms as:
\[ \frac{1}{\log_a x} = \log a, \quad \frac{1}{\log_b x} = \log b, \quad \frac{1}{\log_c x} = \log c \]
(Note: We are considering \( \log x \) as a common factor and can ignore it for the purpose of establishing the relationship.)

4. **Applying the GP Condition**:
From the GP condition \( b^2 = ac \), we take logarithms:
\[ \log b^2 = \log(ac) \]
Using logarithmic properties:
\[ 2 \log b = \log a + \log c \]

5. **Rearranging the Equation**:
Rearranging gives:
\[ 2 \log b = \log a + \log c \]
This can be interpreted as:
\[ \log b = \frac{1}{2} (\log a + \log c) \]
This shows that \( \log a, \log b, \log c \) are in arithmetic progression.

6. **Conclusion**:
Since \( \log a, \log b, \log c \) are in AP, it follows that:
\[ \frac{1}{\log_a x}, \frac{1}{\log_b x}, \frac{1}{\log_c x} \]
are also in AP.

### Final Answer:
Thus, \( \frac{1}{\log_a x}, \frac{1}{\log_b x}, \frac{1}{\log_c x} \) are in **Arithmetic Progression (AP)**.

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