Consider two arithmetic progressions, $\langle a_n\rangle_{n \in N}$ and $\langle b_n\rangle_{n \in N}$, such that $$\frac{\sum_{r=1}^n a_r}{\sum_{r=1}^n b_r} = \frac{3n+1}{4n+2}$$ Find the ratio of the $n$th terms of both sequences.
I find this is a common problem and has some solutions available as well. I solved it via other methods but have some questions in mind regarding the different solution I share below.
My work:
I first assumed that the sum of the first $n$ terms of the sequence $a_n$ is $3nk + k$, and that for $b_n$ is $4nk+2k$. This means that the first terms of the two sequences are respectively $4k$ and $6k$ and the second terms respectively being $3k$ and $4k$. Hence the ratio in question becomes $\frac{n-5}{2n-8}$.
However, this is incorrect.
Instead if we take
$$\frac{\sum_{r=1}^n a_r}{\sum_{r=1}^n b_r} = \frac{3n^2+n}{4n^2+2n}$$
and hence we let the sum of the progressions be $3n^2k+nk$ and $4n^2k+2nk$, the first two terms of $a_n$ work out to be $4k,10k$ and that of $b_n$ to be $6k,14k$, giving us the correct ratio $\frac{3n-1}{4n-1}$.
Why does adding the extra factor of $n$ in the sum of first $n$ terms helps us find the correct ratio, but not if we cancel it out from the fraction for being a common factor? I think it might be because the sum of $n$ terms of an arithmetic progression is a quadratic in $n$, but I cant fully justify it myself.