<s>[INST] Find the value of $a_4-a_2$ 
Let $a_1<a_2<a_3<a_4$ be positive integers such that $\sum_{i=1}^{4}\frac{1}{a_i}=\frac{11}{6}$. Find the value of $a_4-a_2.$

I do not know how to proceed. I have tried to simplify the summation but did not obtain anything useful. Please help.
 [/INST] $\dfrac{11}{6}=\dfrac{1}{a_1}+\dfrac{1}{a_2}+\dfrac{1}{a_3}+\dfrac{1}{a_4}
\nleq\dfrac12+\dfrac13+\dfrac14+\dfrac15=\dfrac{77}{60} \quad \Longrightarrow \quad a_1=1$
$\dfrac{5}{6}=\dfrac{1}{a_2}+\dfrac{1}{a_3}+\dfrac{1}{a_4}
\nleq\dfrac13+\dfrac14+\dfrac15=\dfrac{47}{60} \quad \Longrightarrow \quad a_2=2$
$\dfrac{1}{3}=\dfrac{1}{a_3}+\dfrac{1}{a_4} \quad \Longrightarrow \quad a_3 a_4=3(a_3+a_4)$
In positive integers solutions of last equation is only $(4,12),(12,4),(6,6)$.
$a_3<a_4 \quad \Longrightarrow \quad (a_3,a_4)=(4,12)$.
$a_4-a_2=10$
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