Let $n$ be a positive integer such that $\displaystyle{\frac{3+4+\cdots+3n}{5+6+\cdots+5n} = \frac{4}{11}}$ Let $n$ be a positive integer such that 
$$\displaystyle{\frac{3+4+\cdots+3n}{5+6+\cdots+5n} = \frac{4}{11}}$$
 then 
$$\displaystyle{\frac{2+3+\cdots+2n}{4+5+\cdots+4n}} = \frac{m}{p}.$$
The question further "Is $m+p$ a prime?"

$k+(k+1)+...+kn=\frac{k}{2}(n+1)(kn-k+1)$ (just the half of the sum of the first and the last terms times the number of the terms)$=\frac{k}{2}(kn^2+n-(k-1))$
Now, from the first equation we get $n$: $$33(3n^2+n-2)=20(5n^2+n-4)$$ $$n^2-13n-14=0$$ $$(n-14)(n+1)=0$$ $$n>0\Rightarrow n=14$$
And from the second equation we get $m$: $$m=n\frac{2n^2+n-1}{8n^2+2n-6}=\frac{189}{53}=3.566...$$
So, $m+n$ is not prime, because it is not even integer...
