If the  ratio of the  sum  of n terms of two  AP's be (7n  + 1):(4n + 27),  then the  ratio of their 11th  terms will be

Let Sn, and S' be the sums of n terms of two AP's and T11 and T11, be the respective 11th term, thenSnSn′=n2[2a+(n−1)d]n22a′+(n−1)d′=7n+14n+27 (given)⇒a+(n−1)2da′+(n−1)2d′=7n+14n+27 Now, put n=21, we get a+10da′+10d′=T11T11′=148111=43