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?"
 A: At first attempt, I was tempted to do $\displaystyle{3+4+\cdots+3n = \frac{3n(3n+1)}{2}-3}$, but there are $4$ expressions of that sort, it is better to find a general form of it like this
$$
\begin{align*}
k+(k+1)+(k+1)+\cdots+kn &= \frac{1}{2} \left[ kn(kn+1)-k(k-1) \right]\\
&= \frac{1}{2}\left[ k(n+1)(kn-k+1)\right] \tag{1}
\end{align*}
$$
Appying $(1)$ to $\displaystyle{\frac{3+4+\cdots+3n}{5+6+\cdots+5n}} = \frac{3(n+1)(3n-3+1)}{5(n+1)(5n-5+1)} =\frac{3(3n-1)}{5(5n-4)} = \frac{4}{11}$, leads us to $\displaystyle{\frac{3n-2}{5n-4}=\frac{20}{33}}$ and further to an expresion $99n-66=100n-80 \Rightarrow n=14$
$$
\displaystyle{\frac{2+3+\cdots+2n}{4+5+\cdots+4n}} =\frac{2(n+1)(2n-2+1)}{4(n+1)(4n-4+1)} = \frac{30\times27}{60\times53} =\frac{27}{106}
$$
$m+p=27+106=133$.  But $133 = 7\times19$.  Therefore the answer is "No, $m+p=133$ is not a prime".
(For those wondering what did I just change, for clarity I changed the right side to be $\frac{m}{p}$, the answer still stays)
A: $k+(k+1)+...+kn=\frac{k}{2}(n+1)(kn-k+1)$ (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...
A: Hint:
$$\sum_{i=1}^ni=\dfrac{n(n+1)}{2}$$

After a bit of painful algebra, $n=14$. And, $m= 27$ and $p=106$, if I assume $(m,p)=1$

