Question

The sum of n terms in two AP 's are in the ratio 3n+1:n+4 then the ratio of 4th terms is ?

Answer 1

given , sum of n terms in two a.ps are in the ratio 3n+1:n+4

let for an a.p let a be the first term and d be the common difference

for another a.p let A be the first term and D be the common difference

then ratio of sum of nterms of this A.p. will be

n/2[2a+(n-1)d]:n/2[2A+(n-1)D] = 3n+1/n+4

2a+(n-1)d:2A+(n-1)D =3n+1/n+4 --------(1)

now, ratio of 4 th terms will be

a+3d:A+3D

multiplying and dividing with 2 we get,

2a+6d:2A+6D

by comparing it with eq(1) ,we get
n=7

now the ratio is 3n+1/n+4
3(7)+1/7+4
22:11
2:1

I hope this will help u :))

Answer 2

Answer:

hope it helps MARK me BRAINLIST and