If a, b, c are in A.P. and x, y, z are in G.P.,

(b) 1

$a, b$ and $c$ are in A.P.

$\therefore 2 b=a+c$       .....(1)

And, $x, y$ and $z$ are in G.P. 

$\therefore y^{2}=x z$

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

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

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

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

$=(\mathrm{xz})^{0}$

$=1$