If `(1)/(sqrt(x-1))+(1)/(sqrt(y-1))+(1)/(sqrt(z-1))gt0andx,y,z,` are in G.P., then `(logx^(2))^(-1),(logxz)^(-1),(logz^(2))^(-1)` are in

Options

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

Explanation

To solve the problem, we need to show that if \( \frac{1}{\sqrt{x-1}} + \frac{1}{\sqrt{y-1}} + \frac{1}{\sqrt{z-1}} > 0 \) and \( x, y, z \) are in geometric progression (G.P.), then \( (\log x^2)^{-1}, (\log xz)^{-1}, (\log z^2)^{-1} \) are in harmonic progression (H.P.).

### Step-by-Step Solution:

1. **Understanding the Condition**:
Given that \( x, y, z \) are in G.P., we can express \( y \) in terms of \( x \) and \( z \):
\[ y = \sqrt{xz} \]

2. **Using the Logarithmic Identity**:
We know that if \( y = \sqrt{xz} \), then:
\[ y^2 = xz \]

3. **Taking Logarithms**:
Taking logarithms on both sides gives:
\[ \log y^2 = \log(xz) \]
Using the properties of logarithms, we can rewrite this as:
\[ 2 \log y = \log x + \log z \]

4. **Rearranging the Equation**:
Rearranging the above equation leads to:
\[ 2 \log y = \log x + \log z \]
This implies that:
\[ 2 \log y = \log x + \log z \implies 2b = a + c \]
where \( a = \log x \), \( b = \log y \), and \( c = \log z \).

5. **Identifying Arithmetic Progression**:
The equation \( 2b = a + c \) indicates that \( \log x, \log y, \log z \) are in arithmetic progression (A.P.).

6. **Converting to Harmonic Progression**:
If \( a, b, c \) are in A.P., then their reciprocals \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \) are in harmonic progression (H.P.). Thus:
\[ \frac{1}{\log x}, \frac{1}{\log y}, \frac{1}{\log z} \text{ are in H.P.} \]

7. **Expressing in Terms of the Given Expression**:
We need to express this in terms of the original question. We have:
\[ \frac{1}{\log x^2}, \frac{1}{\log(xz)}, \frac{1}{\log z^2} \]
Since \( \log(xz) = \log x + \log z \), we can see that:
\[ \frac{1}{\log x^2}, \frac{1}{\log(xz)}, \frac{1}{\log z^2} \text{ are in H.P.} \]

### Conclusion:
Thus, we conclude that \( (\log x^2)^{-1}, (\log xz)^{-1}, (\log z^2)^{-1} \) are in harmonic progression.