Let Sn denote the sum of the first n terms of an A.P. if S2n=3Sn then S3n:Ss=

S2nSn=31 so, 2n2[2a+(2n−1)d]n2[2a+(n−1)d]=314a+(4n−2)d2a+(n−1)d=34a+(4n−2)d=6a+(3n−3)d(n+1)d=2aS3nSn=3n2[2a+(3n−1)d]n2[2a+(n−1)d]=3[(n+1)+(3n−1)][(n+1)+(n−1)]=3(4n)2n=6