The sum of n terms of two AP are in the ratio (2n+3):(6n+5),then the ratio of their 13th terms is
Please answer with complete steps and explanation.
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