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. 

Answer

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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
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