If a, b c are in GP and `a^((1)/(x)) =b^((1)/(y)) =c^((1)/(z))`, then x, y, z 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) and \( a^{\frac{1}{x}} = b^{\frac{1}{y}} = c^{\frac{1}{z}} \), then \( x, y, z \) are in arithmetic progression (AP).

### Step-by-Step Solution:

1. **Understanding GP**:
Since \( a, b, c \) are in GP, we can express \( b \) in terms of \( a \) and \( c \):
\[ b = \sqrt{ac} \]

2. **Setting up the equation**:
Given that \( a^{\frac{1}{x}} = b^{\frac{1}{y}} = c^{\frac{1}{z}} \), let's denote this common value as \( k \):
\[ a^{\frac{1}{x}} = k, \quad b^{\frac{1}{y}} = k, \quad c^{\frac{1}{z}} = k \]

3. **Expressing \( a, b, c \) in terms of \( k \)**:
From the equations above, we can express \( a, b, c \) as:
\[ a = k^x, \quad b = k^y, \quad c = k^z \]

4. **Substituting \( b \) in terms of \( a \) and \( c \)**:
Since \( b = \sqrt{ac} \), we substitute the expressions for \( a, b, c \):
\[ k^y = \sqrt{k^x \cdot k^z} \]
Simplifying the right side:
\[ k^y = \sqrt{k^{x+z}} = k^{\frac{x+z}{2}} \]

5. **Equating the exponents**:
Since the bases are the same (both are \( k \)), we can equate the exponents:
\[ y = \frac{x + z}{2} \]

6. **Rearranging the equation**:
Rearranging gives us:
\[ 2y = x + z \]

7. **Conclusion**:
The equation \( 2y = x + z \) indicates that \( x, y, z \) are in arithmetic progression (AP).

### Final Result:
Thus, we conclude that if \( a, b, c \) are in GP and \( a^{\frac{1}{x}} = b^{\frac{1}{y}} = c^{\frac{1}{z}} \), then \( x, y, z \) are in AP.