解设{an}的公比q,S1=a1.S2=a1(1+q).S3=a1(1+q+q^2)
由S1.2S2.3S3诚等差数列,得
S1+3S3=4S2.
a1+3a1(1+q+q^2)=4 a1(1+q),a1≠0得
1+3(1+q+q^2)=4 (1+q)
4+3q+3q^2=4+4q
3q^2-q=0
(3q-1)q=0
q=0,（舍去）,故q=1/3,公比为1/3.