思路分析:要证x,y,z成等比数列,需由条件推出y2=xz.
证法一:∵a,b,c成等差数列,∴b=
.
又∵(b-c)logmx+(c-a)logmy+(a-b)·logmz=0,
∴(
-c)logmx+(c-a)logmy+(a-
)logmz=0.
∴(a-c)(logmx-2logmy+logmz)=0.
∵a,b,c的公差不为零,∴a-c≠0.
∴logmx-2logmy+logmz=0.
∴logmy2=logmxz.
∴y2=xz.
又由题设知,x,y,z均不为零,
∴x,y,z成等比数列.
证法二:设等差数列a,b,c的公差为d(d≠0),则b-c=-d,c-a=2d,a-b=-d,代入已知条件式,得-d(logax-2logmy+logmz)=0.
∵d≠0,∴logmx-2logmy+logmz=0.
∴logmy2=logmxz.
∴y2=xz.
∵由题设知x,y,z均不为0,
∴x,y,z成等比数列.