Question

If a, b, c are in A.P. and x. y, z are in G.P., then the value of $x^{b-c}{y}^{c-a}{z}^{a-b}$ is

• A. 0
• B. 1
• C. xyz
• D. $x^a$ $y^b$ $z^c$

Answer 1

The correct option is B

1

(b) (1)

a, b and c are in A.P.

$\therefore 2b=a+c$ ........ (i)

And x, y and z are in G.P.

$\therefore y^2=xz$

Now, $x^{b-c}{y}^{c-a}{z}^{a-b}$

$=x^{b+a-2b}{y}^{2b-a-a}{z}^{a-b}$ [From (i)]

$=x^{a-b}{y}^{2(a-b)}{z}^{a-b}$

$=\left(xz{)}^{a-b}\right(xz{)}^{b-a}$ [From (ii), $y^2=xz$ ]

$=(xz{)}^0$

$=1$