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. 
 A: If what you call $k$ is independent of $n$, then your argument works fine. But there is no reason to assume independence.
A: $${S_n\over3n+8}={s_n\over7n+15},$$
but we cannot know that
$${S_n\over3n+8}={s_n\over7n+15}\overset{?}{=}{S_{n-1}\over3(n-1)+8}={s_{n-1}\over7(n-1)+15}.$$
Actually,
$${S_n\over3n+8}={s_n\over7n+15}=k_\color{red}{n}.$$
A: Hint.
We have that
$$
S_n=A_1+\cdots+A_n,
$$
where $A_n=Kn+L$, and hence
$$
S_n=K\frac{n(n+1)}{2}+Ln.
$$
Similarly, $s_n=a_1+\cdots+a_n$, where $a_n=kn+\ell$ and $s_n=k\frac{n(n+1)}{2}+\ell n$,
and
$$
\frac{S_n}{s_n}=\frac{K\frac{n(n+1)}{2}+Ln}{k\frac{n(n+1)}{2}+\ell n}=\frac{K(n+1)+2L}{k(n+1)+2\ell}=\frac{Kn+2L+K}{kn+2\ell+k}=\frac{3n+8}{7n+15},
$$
and
$$
\frac{A_{12}}{B_{12}}=\frac{12K+L}{12k+\ell}
$$
A: 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.
A: A short and direct answer to your question, distilling the wisdom in answers given by others here earlier:
Both $S_n$ and $s_n$ are of the form $n(An+B)$. 
However when you take the ratio, $n$ gets cancelled out. 
Therefore when you take the ratio of the $\underline{\text{difference}}$, $n$ (and also $n-1$) should be reinstated accordingly to arrive at the correct answer.
In summary,
$$\begin{align}
\dfrac{S_n}{s_n}&=\dfrac{3n+8}{7n+15}=\dfrac{n(3n+8)}{n(7n+15)}\\
\dfrac{S_{12}}{s_{12}}&=\dfrac {44}{99}\qquad =\dfrac {12\cdot 44}{12\cdot 49}\\
\dfrac{S_{12}-S_{11}}{s_{12}-s_{11}}&\quad\qquad\;\quad=\dfrac{12\cdot44-11\cdot41}{12\cdot 99-11\cdot 92}=\dfrac{48-41}{108-92}=\dfrac7{16}\\
\end{align}$$
