Given that, a, b, c are in A.P. and x, y, z are in G.P. Let d be the common difference between A.P. consecutive terms. So, b−a=c−b=d,c−a=2d Now, xb−c⋅yc−a⋅za−b =x−d⋅y2d⋅z−d=(xz)−d⋅(y2)−d Since, xz=y2 then (xz)−d(xz)d=1