The sum of n terms of two AP are in the ratio (2n+3):(6n+5),then the ratio of their 13th terms is 
(a) 53:155
(b)29:83
(c) 31:89
(d)27:8
Please answer with complete steps and explanation. 

Hi Diksha Raturi dear, option a i.e 53 : 155 is the correct one
Solution:
Given n/2 * [ 2a₁ + (n-1) d₁ ] / ​n/2 * [ 2a₂ + (n-1) d₂ ] = (2n + 3)/(6n + 5)
Cancelling n/2 we have 
 [ 2a₁ + (n-1) d₁ ] /  [ 2a₂ + (n-1) d₂ ] = (2n + 3)/(6n + 5) ----(1)
Now needed is (a₁ + 12 d₁) / (a₂ + 12 d₂)
If we substitute n = 25 then we get ​(a₁ + 12 d₁) / (a₂ + 12 d₂) after cancelling common 2 both in numerator and denominator
So right side would give 50 + 3 / 150 + 5 = 53/155