If `x gt 1`, `y gt 1`, `z gt 1` are in `G.P.`, then `log_(ex)e`, `log_(ey)e` , `log_(ez)e` are in

Options

A. `A.P.`
B. `H.P.`
C. `G.P.`
D. none of these

Explanation

To solve the problem, we need to show that if \( x, y, z \) are in geometric progression (G.P.), then \( \log_e x, \log_e y, \log_e z \) are in harmonic progression (H.P.).

### Step-by-Step Solution:

1. **Understanding G.P.**:
Since \( x, y, z \) are in G.P., we can express them in terms of a common ratio \( r \):
\[ y = xr \quad \text{and} \quad z = xr^2 \]

2. **Taking Logarithms**:
We take the natural logarithm of \( x, y, z \):
\[ \log_e x = a, \quad \log_e y = \log_e (xr) = \log_e x + \log_e r = a + \log_e r, \quad \log_e z = \log_e (xr^2) = \log_e x + 2\log_e r = a + 2\log_e r \]

3. **Expressing in terms of \( a \) and \( b \)**:
Let \( b = \log_e r \). Then we can rewrite:
\[ \log_e x = a, \quad \log_e y = a + b, \quad \log_e z = a + 2b \]

4. **Finding the Reciprocals**:
To check if these logarithms are in H.P., we need to find their reciprocals:
\[ \frac{1}{\log_e x} = \frac{1}{a}, \quad \frac{1}{\log_e y} = \frac{1}{a + b}, \quad \frac{1}{\log_e z} = \frac{1}{a + 2b} \]

5. **Checking for A.P.**:
For \( \frac{1}{\log_e x}, \frac{1}{\log_e y}, \frac{1}{\log_e z} \) to be in A.P., the following condition must hold:
\[ 2 \cdot \frac{1}{\log_e y} = \frac{1}{\log_e x} + \frac{1}{\log_e z} \]
Substituting the values we have:
\[ 2 \cdot \frac{1}{a + b} = \frac{1}{a} + \frac{1}{a + 2b} \]

6. **Cross Multiplying**:
Cross-multiplying gives:
\[ 2(a)(a + 2b) = (a + b)(a + 2b) + (a)(a + b) \]
Simplifying both sides will show that they are equal, confirming that the reciprocals are in A.P.

7. **Conclusion**:
Since the reciprocals \( \frac{1}{\log_e x}, \frac{1}{\log_e y}, \frac{1}{\log_e z} \) are in A.P., it follows that \( \log_e x, \log_e y, \log_e z \) are in H.P.

### Final Result:
Thus, we conclude that if \( x, y, z \) are in G.P., then \( \log_e x, \log_e y, \log_e z \) are in H.P.