Find if terms are terms of the same arithmetic progression 
Is it possible that numbers $\frac{1}{2}, \frac{1}{3}, \frac{1}{5}$ are (not necessarily adjacent) terms of the same arithmetic progression?
  Hint: Yes. Try $\frac{1}{30}$ as a difference.

I  was going back and forth how they found out that difference.
My idea was since we have an arithmetic sequence defined as $a, a+d, a+2d,...$ I thought I could solve for the difference $d=\frac{1}{30}$. Since:
$$\frac{1}{3} = \frac{1}{2}+nd$$
And
$$\frac{1}{5} = \frac{1}{3}+md$$
Then $nd = \frac{1}{3} - \frac{1}{2} = -\frac{1}{6}$ and $md = \frac{1}{5} - \frac{1}{3} = -\frac{2}{15}$
Since it is also part of the same sequence we can find: 
$$nd + md = -\frac{1}{6} -\frac{2}{15} = -\frac{3}{10}$$
Now I'm stuck since I can't see how this brings me any closer to find $m, n, d$. 

Let the terms be $a_1=\frac12, a_m=\frac13,a_n=\frac15$. Then:
$$d=\frac{a_m-a_1}{m-1}=\frac{a_n-a_1}{n-1} \Rightarrow \\
d=\frac{\frac13-\frac12}{m-1}=\frac{\frac15-\frac12}{n-1} \Rightarrow \\
-\frac16(n-1)=-\frac3{10}(m-1) \Rightarrow \\
n=\frac{9}{5}(m-1)$$
As $m,n\in \mathbb N_+$ and $n>m$, we get $m=5k+1,n=9k,d=-\frac1{30k},k\in\mathbb N_+$:
$$\begin{array}{c|c|c|c|c}
m&6&11&16&\cdots\\
\hline
n&9&18&27&\cdots\\
\hline
d&-\frac1{30}&-\frac1{60}&-\frac1{90}&\cdots
\end{array}$$
Note: It was considered a decreasing AP. It can also be considered an increasing AP.
