Arithmetic Progression. Q. The ratio between the sum of $n$ terms of two A.P's is $3n+8:7n+15$. Find the ratio between their $12$th term.
My method:
Given:
$\frac{S_n}{s_n}=\frac{3n+8}{7n+15}$
$\frac{S_n}{3n+8}=\frac{s_n}{7n+15}=k$
$\frac{T_n}{t_n}=\frac{S_n-S_{n-1}}{s_n-s_{n-1}}=\frac{k\left(\left(3n+8\right)-\left(3\left(n-1\right)+8\right)\right)}{k\left(\left(7n+15\right)-\left(7\left(n-1\right)+15\right)\right)}=\frac{3}{7}$
As this applies for any term:
$\frac{T_{12}}{t_{12}}=\frac{3}{7}$
But this is not the answer. The actual answer is $\frac7{16}$. I know how to obtain that answer.
But why is my solution wrong? Its probably the concept I guess. 

The trick is to notice that we can safely cancel constant common factors (if any), and stick in the $n$  in both numerator and denominator:
$\dfrac{T_n}{t_n} = \dfrac{S_n - S_{n - 1}}{s_n - s_{n - 1}} = \dfrac{n(3n + 8) - (n - 1)(3(n - 1) + 8)}{n(7n + 15) - (n - 1)(7(n - 1) + 15)} = \dfrac{6n + 5}{14n + 8}$
Plug in $n = 12$ to get the ratio of 12th terms.
