If d be the common difference of A.P., then
logyx=1+d,logzy=1+2d,−15logxz=1+3d
∴x=y1+d,y=z1+2d,z=x−1+3d15
Eliminating y and z, we get
x=z(1+d)(1+2d)=x−(1−3d)(1+d)(1+2d)15
∴(1+d)(1+2d)(1+3d)=−15
or 6d3+11d2+6d+16=0
(d+2)(6d2−d+8)=0
∴d=−2
The other factor gives imaginary values of d.
∴x=y−1,y=z−3,z=x1/3
or x=y−1=z3,y=z−3.