Showing that a progression is arithmetic

this one is from Gelfand's book "Algebra".

Problem 204. Is it possible that numbers $\frac{1}{2}$, $\frac{1}{3}$, and $\frac{1}{5}$ are (not necessarily adjacent) terms of the same arithmetic progression?

Is there a method to show if they're from same progression? or should I just try different differences?

All that came to my mind was to write system of equations: $$\left\{\begin{array}\frac{1}{2}-nd=\frac{1}{3}\\\frac{1}{3}-kd=\frac{1}{5}\end{array}\right.$$ But it can't be solved for $d$ (difference).

By the way, answer is $d=-\frac{1}{30}$, which is $-2*5*3$, so maybe the difference depends on denominators of progression?

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Multiplying by the common denominator, our three terms are $15,10,6$. These are (non-adjacent) terms in the sequence $15,14,13,12,11,10,9,8,7,6$. Then divide them all by that same common denominator.