If `a^x=b^y=c^z` and `a,b,c` are in G.P. show that `1/x,1/y,1/z` are in A.P.

Answer

To solve the problem, we need to show that if \( a^x = b^y = c^z \) and \( a, b, c \) are in geometric progression (G.P.), then \( \frac{1}{x}, \frac{1}{y}, \frac{1}{z} \) are in arithmetic progression (A.P.).

### Step-by-Step Solution:

1. **Set the Common Value**:
Since \( a^x = b^y = c^z \), we can denote this common value as \( k \). Therefore, we have:
\[ a^x = k, \quad b^y = k, \quad c^z = k \]

2. **Express \( a, b, c \) in terms of \( k \)**:
From the equations above, we can express \( a, b, c \) as:
\[ a = k^{\frac{1}{x}}, \quad b = k^{\frac{1}{y}}, \quad c = k^{\frac{1}{z}} \]

3. **Use the G.P. condition**:
Since \( a, b, c \) are in G.P., we have the relationship:
\[ b^2 = ac \]
Substituting the expressions for \( a, b, c \):
\[ (k^{\frac{1}{y}})^2 = k^{\frac{1}{x}} \cdot k^{\frac{1}{z}} \]

4. **Simplify the equation**:
This simplifies to:
\[ k^{\frac{2}{y}} = k^{\frac{1}{x} + \frac{1}{z}} \]

5. **Equate the exponents**:
Since the bases are the same (both are \( k \)), we can equate the exponents:
\[ \frac{2}{y} = \frac{1}{x} + \frac{1}{z} \]

6. **Rearranging the equation**:
To show that \( \frac{1}{x}, \frac{1}{y}, \frac{1}{z} \) are in A.P., we can rearrange the equation:
\[ \frac{1}{x} + \frac{1}{z} = \frac{2}{y} \]
This can be rewritten as:
\[ \frac{1}{x} - \frac{2}{y} + \frac{1}{z} = 0 \]

7. **Conclusion**:
The equation \( \frac{1}{x} - \frac{2}{y} + \frac{1}{z} = 0 \) indicates that \( \frac{1}{x}, \frac{1}{y}, \frac{1}{z} \) are in A.P. Hence, we have proved the statement.