∵an+1=3Sn,∴Sn+1-Sn=3Sn,∴Sn+1=4Sn,若S1=0,则数列{an}为等差数列;若S1≠0,则数列{Sn}为首项为S1,公比为4的等比数列,∴Sn=S1?4n-1,此时an=Sn-Sn-1=3S1?4n-2(n≥2),即数列从第二项起,后面的项组成等比数列.综上,数列{an}可能为等差数列,但不会为等比数列.故选C.