The ratio of their $n$-th term.

It is actually quite simple. Let $a_1$ and $a_1'$ denote the first terms of the first and second progressions with their common differences $d$ and $d'$ respectively. We thus get $$\frac{S_1}{S_2} = \frac {0.5n (2a_1 +(n-1)d)}{0.5n (2a_1' +(n-1)d')} = \frac {2a_1+(n-1)d}{2a_1' +(n-1)d'} = \frac {7n+1}{4n+27} $$

The ratio of the $n$th term of the two AP's can be thus calculated as $$\frac{a_n}{a_n'} = \frac {a_1 +(n-1)d}{a_1'+(n-1)d'} = \frac {2a_1 +((2n-1)-1)d}{2a_1' + ((2n-1)-1)d'} $$ $$=\frac {S_{2n-1}}{S_{2n-1}'} = \frac {14n-6}{8n+23} $$ Hope it helps.

Hint: The sum of $nth$ term of a AP is given by $$\frac{n}{2}[a+a_n]$$.The ratio of first term can easily be found out by putting $n=1$.So it is easy to find out the ratio of $nth$ term using this.