If a, b and c are in GP, then `log_(a)x, log_(b)x and log_(c) x` will be

Options

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

Answer

To solve the problem, we need to show that if \( a, b, c \) are in geometric progression (GP), then \( \log_a x, \log_b x, \log_c x \) will be in arithmetic progression (AP).

### Step-by-Step Solution:

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

2. **Expressing Logarithms**:
We can express the logarithms in terms of a common base, say base \( x \):
\[ \log_a x = \frac{\log x}{\log a}, \quad \log_b x = \frac{\log x}{\log b}, \quad \log_c x = \frac{\log x}{\log c} \]

3. **Finding the Condition for AP**:
For three numbers \( p, q, r \) to be in AP, the following condition must hold:
\[ 2q = p + r \]
In our case, substituting \( p = \log_a x \), \( q = \log_b x \), and \( r = \log_c x \):
\[ 2 \log_b x = \log_a x + \log_c x \]

4. **Substituting the Logarithmic Expressions**:
Substituting the expressions from step 2 into the AP condition:
\[ 2 \cdot \frac{\log x}{\log b} = \frac{\log x}{\log a} + \frac{\log x}{\log c} \]

5. **Simplifying the Equation**:
We can factor out \( \log x \) (assuming \( \log x \neq 0 \)):
\[ 2 \cdot \frac{1}{\log b} = \frac{1}{\log a} + \frac{1}{\log c} \]

6. **Cross-Multiplying**:
Cross-multiplying gives us:
\[ 2 \log a \log c = \log b (\log a + \log c) \]

7. **Using the GP Condition**:
Recall from step 1 that \( b^2 = ac \) implies:
\[ \log b^2 = \log(ac) \implies 2 \log b = \log a + \log c \]
This confirms that:
\[ 2 \cdot \frac{1}{\log b} = \frac{1}{\log a} + \frac{1}{\log c} \]
holds true, thus proving that \( \log_a x, \log_b x, \log_c x \) are in AP.

### Conclusion:
Therefore, if \( a, b, c \) are in GP, then \( \log_a x, \log_b x, \log_c x \) are in AP.