Mathematics Question from progression

The sum of n terms of two artimetic series are in the ratio of (7n+1):(4n+27). Find the ratio of their nth term.
Answer
(14n-6):(8n+23)
please show me the steps of solving it.
Thanx.

Dear, Do like this,
You may be aware sum of arithmetic series is of the form
Sn = 2a+(n-1)d. Therefore for series 1 sn1 is the sum and sn2 is the sum for series 2.
Now Sn1/Sn2 = 7n+1/4n+27 = R
Now sum adjustment of numbers so that to bring in the form of 2a+(n-1)d.
R = (7n+1)/(4n+27) = (1+7(n-1+1))/(27+4(n-1+1)) = (1+7(n-1)+7)/(27+4(n-1)+4)
R = (8+7(n-1))/(31+4(n-1)). Now in the denominator the number 31 is odd number. Hence to get a even number multiply 2 both in numerator and denominator. Therefore

R= 2*(8+7(n-1))/2*(31+4(n-1))
= (2*(8)+14(n-1))/(2*(31)+8(n-1)). Now by inspection of numerator and denominator we can see it is of the form 2a+(n-1)d. Hence for series1 a=8, d=14 for series2 a=31, d= 8. Therefore nth terms for series 1 and 2 can be written as tn1= 8+(n-1)14, tn2= 31+(n-1)8.

Therefore the ratio of nth terms is as follows.
tn1/tn2= (8+(n-1)14)/(31+(n-1)8) = (8+14n-14)/(31+8n-8) = (14n-6)/(8n+23). Which is the desired answer.

[n/2*(2a1+(n-1)d1)]/[n/2*(2a2+(n-1)d2] = (7n+1)/(4n+2)
comparing the right hand side and the left hand side,
2a1+(n-1)d1 = 7n+1
2a2+(n-1)d2 = 4n+2
so..
a1=4
a2=31/2
d1=7
d2=4
therefore..
ration of the nth terms are..
[a1+(n-1)d1]/[a2+(n-1)d2]
substitute the values.. you will get the answer.