Question

The sum of n terms of two APs are in ratio 7n+1/4n+27.

Find the ratio of their 11th terms

Answer 1

Answer.
Given ratio of sum of n terms of two AP’s = (7n+1):(4n+27) Let’s consider the ratio these two AP’s mth terms as am : a’m →(2) Recall the nth term of AP formula, an = a + (n – 1)dHence equation (2) becomes, am : a’m = a + (m – 1)d : a’ + (m – 1)d’ On multiplying by 2, we get am : a’m = [2a + 2(m – 1)d] : [2a’ + 2(m – 1)d’] = [2a + {(2m – 1) – 1}d] : [2a’ + {(2m – 1) – 1}d’] = S2m – 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].
so ratio of 11th terms is
14*11-6/8*11+23
that is 154-6/88+23
=148/111