Question

The ratio between the sum of n terms of two A.P.'s is (7n+1):(4n+27). Find the ratio of their 11th terms.

Answer 1

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)d
Hence 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].

m=11
so the ratio of 11th terms of two AP’s is [14*11– 6] : [8*11 + 23] = 148: 111