The ratio of sums of n terms of two A.P'. is `(7n + 1) / (4n + 27)`. Find the ratio of their `11^(th)` terms.

Explanation

Let `a_(1) and a_(2)` be the first terms and `d_(1) and d_(2)` be the common differences of two A.P.s respectively then
`((n)/(2) [2a_(1) + (n - 1) d_(1)])/((n)/(2) [2a_(2) + (n - 1) d_(2)]) = (7n + 1)/(4n + 27) rArr (a_(1) + ((n - 1)/(2))d_(1))/(a_(2) + ((n - 1)/(2)) d_(2)) = (7n + 1)/(4n + 27)`
For ratio of `11^(th)` terms
`(n - 1)/(2) = 10 rArr n = 21`
so ratio of `11^(th)` terms is `(7(21) + 1)/(4(21) + 27) = (148)/(111) = (4)/(3)`