If `a^(x) = b^(y) = c^(z)` and a, b, c are in G.P. then x, y, z are in

Options

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

Answer

To solve the problem, we start with the given conditions:

1. \( a^x = b^y = c^z \)
2. \( a, b, c \) are in Geometric Progression (G.P.).

### Step 1: Express \( a, b, c \) in terms of a common ratio
Since \( a, b, c \) are in G.P., we can express them as:
- \( b = ar \)
- \( c = ar^2 \)

where \( r \) is the common ratio.

### Step 2: Set the common value
Let \( k = a^x = b^y = c^z \). Then we can express \( a, b, c \) in terms of \( k \):
- From \( a^x = k \), we get \( a = k^{1/x} \)
- From \( b^y = k \), we get \( b = k^{1/y} \)
- From \( c^z = k \), we get \( c = k^{1/z} \)

### Step 3: Substitute \( b \) and \( c \) in terms of \( a \)
Using the expressions for \( b \) and \( c \):
- \( b = ar \) implies \( k^{1/y} = k^{1/x} r \)
- \( c = ar^2 \) implies \( k^{1/z} = k^{1/x} r^2 \)

### Step 4: Relate the expressions
From the first equation:
\[ k^{1/y} = k^{1/x} r \implies r = \frac{k^{1/y}}{k^{1/x}} = k^{(1/y - 1/x)} \]

From the second equation:
\[ k^{1/z} = k^{1/x} r^2 \implies r^2 = \frac{k^{1/z}}{k^{1/x}} = k^{(1/z - 1/x)} \]

### Step 5: Equate the expressions for \( r \)
Now we have two expressions for \( r \):
1. \( r = k^{(1/y - 1/x)} \)
2. \( r^2 = k^{(1/z - 1/x)} \)

Substituting the first expression into the second:
\[ (k^{(1/y - 1/x)})^2 = k^{(1/z - 1/x)} \]
This simplifies to:
\[ k^{2(1/y - 1/x)} = k^{(1/z - 1/x)} \]

### Step 6: Set the exponents equal
Since the bases are the same, we can set the exponents equal:
\[ 2\left(\frac{1}{y} - \frac{1}{x}\right) = \frac{1}{z} - \frac{1}{x} \]

### Step 7: Rearranging the equation
Rearranging gives:
\[ 2\frac{1}{y} - 2\frac{1}{x} = \frac{1}{z} - \frac{1}{x} \]
\[ 2\frac{1}{y} = \frac{1}{z} + \frac{1}{x} \]

### Step 8: Conclusion
This shows that \( x, y, z \) are in Harmonic Progression (H.P.) since the relationship we derived is characteristic of H.P.

### Final Answer
Thus, \( x, y, z \) are in Harmonic Progression.

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