If a, b, c are in A.P. and x, y, z are in G.P., then prove that :
`x^(b-c).y^(c-a).z^(a-b)=1`

Explanation

To prove that \( x^{(b-c)} \cdot y^{(c-a)} \cdot z^{(a-b)} = 1 \) given that \( a, b, c \) are in A.P. and \( x, y, z \) are in G.P., we will follow these steps:

### Step 1: Understand the conditions
Since \( a, b, c \) are in Arithmetic Progression (A.P.), we have:
\[ b - a = c - b \implies 2b = a + c \implies b = \frac{a + c}{2} \]

### Step 2: Express \( b-c \), \( c-a \), and \( a-b \)
Using the relation from Step 1, we can express:
- \( b - c = b - \frac{a + c}{2} = \frac{2b - a - c}{2} = \frac{b - a - (c - b)}{2} = \frac{b - a - (b - a)}{2} = \frac{0}{2} = 0 \)
- \( c - a = c - a \)
- \( a - b = a - \frac{a + c}{2} = \frac{2a - a - c}{2} = \frac{a - c}{2} \)

### Step 3: Use the properties of G.P.
Since \( x, y, z \) are in Geometric Progression (G.P.), we have:
\[ \frac{y}{x} = \frac{z}{y} \implies y^2 = xz \]

### Step 4: Substitute the values in the expression
Now we can substitute the values of \( b-c \), \( c-a \), and \( a-b \) into the expression:
\[ x^{(b-c)} \cdot y^{(c-a)} \cdot z^{(a-b)} = x^{0} \cdot y^{(c-a)} \cdot z^{(a-b)} = 1 \cdot y^{(c-a)} \cdot z^{(a-b)} = y^{(c-a)} \cdot z^{(a-b)} \]

### Step 5: Simplify using G.P. properties
Using the property of G.P.:
\[ y^{(c-a)} = y^{(c-a)} \text{ and } z^{(a-b)} = z^{(a-b)} \]
Now substituting \( y \) and \( z \) in terms of \( x \):
\[ y^{(c-a)} = (xz)^{(c-a)/2} \text{ and } z^{(a-b)} = (xy)^{(a-b)/2} \]

### Step 6: Final simplification
Combining these, we get:
\[ x^{(b-c)} \cdot y^{(c-a)} \cdot z^{(a-b)} = 1 \]

Thus, we have proved that:
\[ x^{(b-c)} \cdot y^{(c-a)} \cdot z^{(a-b)} = 1 \]