If $$x, y$$ and $$z$$ are $${p^{{\text{th}}}},{q^{{\text{th}}}}\,{\text{and }}{r^{{\text{th}}}}$$   terms respectively of an A.P. and also of a G.P., then $${x^{y - z}}{y^{z - x}}{z^{x - y}}$$   is equal to:

Solution :
$$\eqalign{ & \because \,\,x,y,z{\text{ are the }}{p^{{\text{th}}}},{q^{{\text{th}}}}{\text{ and }}{r^{{\text{th}}}}\,{\text{terms of an A}}{\text{.P}}{\text{.}} \cr & \therefore \,\,\,x = A + \left( {p - 1} \right)D;y = A + \left( {q - 1} \right)D;z = A + \left( {r - 1} \right)D \cr & \Rightarrow \,\,x - y = \left( {p - q} \right)D;y - z = \left( {q - r} \right)D;z - x = \left( {r - p} \right)D\,\,\,\,\,\,\,.....\left( 1 \right) \cr & {\text{where }}A{\text{ is the first term and }}D{\text{ is the common difference}}{\text{.}} \cr & {\text{Also }}x,y,z{\text{ are the }}{p^{{\text{th}}}},{q^{{\text{th}}}}{\text{ and }}{r^{{\text{th}}}}\,{\text{terms of a G}}{\text{.P}}{\text{.}} \cr & \therefore \,\,x = A{R^{p - 1}},y = A{R^{q - 1}},z = A{R^{r - 1}} \cr & \therefore \,\,{x^{y - z}}{y^{z - x}}{z^{x - y}} = {\left( {A{R^{p - 1}}} \right)^{y - z}}{\left( {A{R^{q - 1}}} \right)^{z - x}}{\left( {A{R^{r - 1}}} \right)^{x - y}} \cr & = {A^{y - z + z - x + x - y}}{R^{\left( {p - 1} \right)\left( {y - z} \right) + \left( {q - 1} \right)\left( {z - x} \right) + \left( {r - 1} \right)\left( {x - y} \right)}} \cr & = {A^0}{R^{\left( {p - 1} \right)\left( {q - r} \right)D + \left( {q - 1} \right)\left( {r - p} \right)D + \left( {r - 1} \right)\left( {p - q} \right)D}} \cr & = {A^0}{R^0} = 1 \cr} $$