Question

The ratio of the
the
sum of n terms
of two A.P's is (7n + 1)(4+27) Find
the ratio of their nth terms .​

Answer 1

Answer:

let

a and d be the first term and common difference of first AP

p and q be the first term and common difference of second AP

(n/2)(2a+(n-1)d) / (n/2)(2p+(n-1)q = (7n+1)(4n+27)

(n is missing in question in denominator. I have assumed it to be 4n+27

(2a+ (n-1)d)/(2p+(n-1)q) = (7n+1)/(4n+27)

(a+(n-1)d/2)/(p+(n-1)q/2) = (7n+1)/(4n+27)......(1)

if we replace (n-1)/2 by n-1 the LHS will become the ratio of nth terms of the APs which we want to find.

so (n'-1)/2 = n-1

n'-1 = 2n-2

n' = 2n-1

so putting 2n-1 in place of n in eqn (1) we have

(a+(n-1)d)/(p+(n-1)q) = (7(2n-1)+1)/(4(2n-1)+27)

= (14n-7+1) /(8n-4+27)

Tn/T'n = (14n - 6)/(8n + 23)