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
2:3
3:4
4:3
5:6
Let Sn, and S' be the sums of n terms of two AP's and T11 and T11, be the respective 11th term, then
SnSn′=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