∵a(b-c)x^2+b(c-a)x+c(a-b)=0有等根
∴Δ＝[b(c-a)]^2-4[a(b-c)][c(a-b)]=0
a^2b^2+b^2c^2-2acb^2
-4bca^2+4acb^2+4a^2c^2-4abc^2=0,
a^2b^2+b^2c^2+2acb^2-4ac(ab+bc)+4a^2c^2=0
(ab+bc)^2-4ac(ab+bc)+4a^2c^2=0
(ab+bc-2ac)^2=0
∴ab+bc-2ac=0,
ab+bc=2ac,两边同除以abc得：(1/c)+(1/a)=2/b,
∴2/b=1/a+1/c
∴1/a,1/b,1/c成等差数列