THE RATIO OF THE SUM OF n terms of two A.P. is(7n+1):(4n+27).find the ratio of 11th terms

Given ratio of sum of n terms of two AP’s = (7n+1):(4n+27)͢   (1)
Let’s consider the ratio these two AP’s mth terms as a_m : a’_m →(2)
Recall the nth term of AP formula, a_n = a + (n – 1)d
Hence equation (2) becomes,
a_m : a’_m = a + (m – 1)d : a’ + (m – 1)d’
On multiplying by 2, we get
a_m : a’_m = [2a + 2(m – 1)d] : [2a’ + 2(m – 1)d’]
= [2a + {(2m – 1) – 1}d] : [2a’ + {(2m – 1) – 1}d’]
= S_{2m – 1} : S’_{2m – 1} 
= [7(2m – 1) + 1] : [4(2m – 1) +27] [from (1)]
= [14m – 7 +1] : [8m – 4 + 27]
= [14m – 6] : [8m + 23]
Thus the ratio of mth terms of two AP’s is [14m – 6] : [8m + 23]
Therefore ,the 11th term ratio=14x11-6/8x11+23=148/111(Answ).