How to find the maximum value of a difference between these values in an arithmetic series? 
The following is an arithmetic progression and $m$ is a positive integer.
$$m, \underbrace{\dots}_{n+1 \text{ terms}}, 33, \underbrace{\dots}_{3n+1 \text{ terms}}, 113$$
Find the maximum value of $n-m$.

Does it require the use of derivatives? I understand that in an arithmetic progression this equation is used:
$$a_{n}=a_{1}+(n-1)d$$
but I don't know how to relate it with the problem or if it would be used at all.
Edit:
Looking at the solution from the source where I found this problem the author addresses it this way:
Because $m>0m$ , $m\in \mathbb{N}$ (sidenote: not sure how to conclude this)
Therefore,
$$m;\underset{\overbrace{\textrm{n+1 terms}}}{\cdots};33;\underset{\overbrace{\textrm{3n+1 terms}}}{\cdots};113$$
33 is positioned in the place labeled as $\textrm{n+3}$, which comes from $\textrm{1+n+1+1}$,
from $a_{n}=a_{1}+(n-1)d$
$a_{n+3}=m+d(n+2)=33$,
then,
$$33;\underset{\overbrace{\textrm{3n+1 terms}}}{\cdots};113$$
$$a_{3n+3}=33+(3n+2)d=113$$
Which leads to:
$$(3n+2)d=80$$
The part which gets tricky comes here as the author concludes that $n=106$, $d=\frac{1}{4}$ (without explaining how).
Replacing $n=106$, $d=\frac{1}{4}$
$$a_{106+3}=m+\frac{1}{4}(106+2)=33$$
$$m=33-\frac{1}{4}(106+2)=6$$
Then the author concludes in order $n-m$ to be maximum $n$ should be maximum and $m$ minimum.
Therefore, $n-m=106-6=100$,
which is the answer provided. Although the above approach does not require calculus topics (i.e derivatives). Instead guesses values based on a lineal equation with two unknowns. This part is something which is confusing. Any help on an alternative that can lead to that answer but with a more logical and easy to follow algorithm?.
 A: $$m+(n+2)d=33\tag1$$
$$m+(4n+4)d=113\tag2$$
Hence we get, from $(1) - (2)$, 
$$(3n+2)d=80 \Rightarrow n=\frac{80}{3d}-\frac{2}{3}$$
And $4\times (1) - (2)$ gives,
$$3m+4d=19 \Rightarrow m=\frac{19}{3}-\frac{4d}{3}$$
So, we get
$$n-m=\frac{80}{3d}-\frac{2}{3}-\frac{19}{3}+\frac{4d}{3}=\frac{4d^2+80}{3d}-7$$
To make $n-m$ maximum, differentiate $\frac{4d^2+80}{3d}$ with respect to $d$ and equate to $0$. Also check the second derivative at that point to ensure it is a maximum.
Now, conclude accordingly.
Hope you can finish this.
A: Letting $d$ be the common difference of successive members, we have
$$m+(n+2)d=33\tag1$$
$$m+(4n+4)d=113\tag2$$
From $(1)(2)$, we have
$$3m=19-4d,\quad 3n+2=\frac{80}{d}$$
Now, since $m$ is a positive integer, we see that $4d$ has to be an integer less than $19$, from which we can write $d=\frac{N}{4}$ where $N\lt 19$ is an integer.
Also, since $n$ is a non-negative integer, we see that $\frac{80}{d}=\frac{320}{N}$ is a positive integer.
It follows from this that $N\lt 19$ has to be a positive divisor of $320$.
So, $N$ has to be either $1,2,4,5,8,10,16$.
Here, using $d=\frac N4$, we have
$$n-m=\frac 13\left(\frac{80}{d}-2\right)-\frac{19-4d}{3}=\frac N3+\frac{320}{3N}-7$$
Now, we define $$f(N)=\frac N3+\frac{320}{3N}-7$$
to have
$$f(N)-f(N+1)=\frac{-N^2-N+320}{3N(N+1)}$$
Here, note that when $f(N)-f(N+1)\gt 0$, i.e. $f(N)\gt f(N+1)$, $f(N)$ is decreasing.
So, we see that $f(N)$ is decreasing when $-N^2-N+320\gt 0\implies (0\lt)\ N\lt \frac{-1+\sqrt{1281}}{2}$ where
$$17=\frac{-1+\sqrt{1225}}{2}\lt \frac{-1+\sqrt{1281}}{2}\lt\frac{-1+\sqrt{1369}}{2}=18$$
Since we already see that $N$ has to be either $1,2,4,5,8,10,16$ (note that all of these values are smaller than $17$), the maximum of $n-m$ is attained when $N$ is the minimum, i.e. $N=1$.
Therefore, the maximum of $n-m$ is $f(1)=100$ when $N=1,d=\frac 14,m=6,n=106$.
A: Just to propose an alternative simple and easy to follow approach (as requested in the OP) that maybe could have been that of who conceived the original problem and did not explain the final step. Starting from $(3n+2)d=80 \,\,\,\,\,$, we have $n=80/(3d) - 2/3\,\,\,\,$. Substituting this in $(n+2)d=33-m\,\,\,\,\,$, we directly get
$$m=\frac{19-4d}{3}$$
Now since $m$ is a positive integer and $d$ must be positive, the only possible values of $d$ are $1/4\,$, $1\,$, and $4\,$. These values lead to $m=6\,$, $m=5\,$, and $m=1\,$, respectively. The corresponding values of $n$ are  $n=106\,$, $n=26\,$, and $n=6\,$, respectively. So the solution that maximizes $n-m\,\,$ is the first one, with $n=106\,\,$ and $d=1/4\,\,$.  
