Question

Ratio of sum of n terms of two AP's is 9n + 1 :2n + 29. Find ratio of their nth terms....
please help me.....
it means a lot... ☺ ​

Answer 1

Answer:

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

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

(n/2)(2a + (n - 1)d). 9n + 1

------------------------- = -----------

(n/2)(2p + (n - 1)q). 2n + 29

(2a + (n - 1)d). 9n + 1

------------------ = -----------

(2p + (n - 1)q). 2n + 29

(a + (n - 1)/2d). 9n + 1

------------------------- = -----------......................(1)

(p + (n - 1)/2q). 2n + 29

now ration of nth terms of the APs would be

a+ (n-1)d

------------

p+ (n-1)q

if we replace n = 2n - 1 in eqn (1), we have

a + (2n -1- 1)/2d 9(2n-1) + 1

------------------------- = -----------

(p + (2n-1 - 1)/2q). 2(2n -1) + 29

a + (2n -2)/2d 18n - 9 + 1

------------------------- = -----------

(p + (2n-2)/2q). 4n - 2 + 29

a + (n -1)/d 18n - 8

------------------------- = -----------

(p + (n-1)/q). 4n + 27