If Sn denotes the sum of first n terms of an A.P. and S3n−Sn−1S2n−S2n−1=31, then the value of n is

S3n=3n2[2a+(3n−1)d]Sn−1=n−12[2a+(n−2)d] ⇒ S3n−Sn−1=12[2a(3n−n+1)]+d2[3n(3n−1)−(n−1)(n−2)]  =122a(2n+1)+d8n2−2=a(2n+1)+d4n2−1=(2n+1)[a+(2n−1)d]S2n−S2n−1=T2n=a+(2n−1)d⇒ S3n−Sn−1S2n−S2n−1=(2n+1)Given,S3n−Sn−1S2n−S2n−1=31⇒n=15