if the ratio of sum of n terms of two APs is (3n+8):(7n+15), then the ratio of 12th terms is??

S_n = n/2(2a + (n-1)d)

(S_n)₁/(S_n)₂ = [2a₁ + (n-1)d₁] / [2a₂ + (n-1)d₂]

                = [a₁ + (n-1)d₁/2 ] / [a₂ + (n-1)d₂/2]

this expression will become ratio of n^th term if , (n-1)/2 is made (k-1) ...

(S_n)₁/(S_n)₂ = 3n+8/7n+15                       (given)

for (n-1)/2 = k-1         or        n = 2k-1

(S_n)₁/(S_n)₂ = (T_n)₁/(T_n)₂ = 3(2k-1) + 8/7(2k-1) + 15 = (6k + 5)/(14k + 8)

now put k=12 , this will give ration of 12^th term

T₁₂ = 77/176

approve my ans if u like it

thnx