Question

If a, b, care in AP and x, y, z are in GP, then the value of X^(b-c) Y^(c-a) Z^(a-b) is

• A. 0
• B. 1
• C. xyz
• D. X^ay^bz^c

Answer 1

The Correct Option is B

We are given that \( a, b, c \) are in arithmetic progression (AP) and \( x, y, z \) are in geometric progression (GP). Since \( a, b, c \) are in AP, we know: \[ 2b = a + c \quad \Rightarrow \quad b = \frac{a + c}{2}. \] Since \( x, y, z \) are in GP, we know: \[ y^2 = xz. \] Now, we need to evaluate the expression \( x^{(b-c)} y^{(c-a)} z^{(a-b)} \). Substituting the value of \( b \) from the AP relation: \[ x^{(b-c)} y^{(c-a)} z^{(a-b)} = x^{\left(\frac{a+c}{2} - c\right)} y^{(c-a)} z^{(a - \frac{a+c}{2})}. \] Simplifying the exponents: \[ x^{\left(\frac{a-c}{2}\right)} y^{(c-a)} z^{\left(\frac{a-c}{2}\right)}. \] Notice that the exponents of \( x \) and \( z \) are equal and opposite, and the exponent of \( y \) is zero since it is the middle term in the progression. Therefore, the expression simplifies to: \[ x^0 y^0 z^0 = 1. \] Thus, the value of the given expression is 1.