If 1,  logyx,logzy,−15 logxz  are in A.P, then

Let d  be the common differences. Then. logyx=1+d⇒x=y1+d             logzy=1+2d⇒y=z1+2dand     −15logxz=1+3d⇒z=x−(1+3d)/15∴             x=y1+d=z(1+2d)(1+d)     =x−(1+d)(1+2d)(1+3d)/15⇒                  (1+d)(1+2d)(1+3d)=−15⇒                  6d3+11d2+6d+16=0⇒                  (d+2)(6d2−d+8)=0⇒d=−2∴                     x=y1+d=y−1,  y=z1−4=z−3and            x=(z−3)−1=z3