1. **Problem statement:** Given that $a, b, c$ are in arithmetic sequence (A.P.) and $x, y, z$ are in geometric sequence (G.P.), prove that $$(x^{b-c})(y^{c-a})(z^{a-b}) = 1.$$ 2. **Recall definitions:** - For $a, b, c$ in A.P., the middle term is the average of the other two: $$b = \frac{a+c}{2}.$$ - For $x, y, z$ in G.P., the middle term squared equals the product of the other two: $$y^2 = xz.$$ 3. **Rewrite the expression:** $$ (x^{b-c})(y^{c-a})(z^{a-b}) = x^{b-c} \cdot y^{c-a} \cdot z^{a-b}. $$ 4. **Substitute $b = \frac{a+c}{2}$:** Calculate each exponent: - $b - c = \frac{a+c}{2} - c = \frac{a - c}{2}$ - $c - a = c - a$ - $a - b = a - \frac{a+c}{2} = \frac{2a - a - c}{2} = \frac{a - c}{2}$ So the expression becomes: $$ x^{\frac{a-c}{2}} \cdot y^{c - a} \cdot z^{\frac{a-c}{2}}. $$ 5. **Rewrite $y^{c-a}$ as $y^{-(a-c)}$:** $$ x^{\frac{a-c}{2}} \cdot y^{-(a-c)} \cdot z^{\frac{a-c}{2}}. $$ 6. **Group terms:** $$ \left(x^{\frac{1}{2}} z^{\frac{1}{2}}\right)^{a-c} \cdot y^{-(a-c)} = \left(\sqrt{xz}\right)^{a-c} \cdot y^{-(a-c)}. $$ 7. **Use the G.P. property $y^2 = xz$:** Taking square root, $$ y = \sqrt{xz}. $$ 8. **Substitute $y = \sqrt{xz}$:** $$ \left(\sqrt{xz}\right)^{a-c} \cdot \left(\sqrt{xz}\right)^{-(a-c)} = 1. $$ 9. **Simplify:** $$ 1. $$ **Final answer:** $$ (x^{b-c})(y^{c-a})(z^{a-b}) = 1. $$ This completes the proof.