<s>[INST] 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. 
 [/INST] 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,
$$\begin{align}
\dfrac{S_n}{s_n}&=\dfrac{3n+8}{7n+15}=\dfrac{n(3n+8)}{n(7n+15)}\\
\dfrac{S_{12}}{s_{12}}&=\dfrac {44}{99}\qquad =\dfrac {12\cdot 44}{12\cdot 49}\\
\dfrac{S_{12}-S_{11}}{s_{12}-s_{11}}&\quad\qquad\;\quad=\dfrac{12\cdot44-11\cdot41}{12\cdot 99-11\cdot 92}=\dfrac{48-41}{108-92}=\dfrac7{16}\\
\end{align}$$
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