The first problem is that you’ve not understood the definition of $A_n$. We’re told that $$A_n=\{5n,5n+1,5n+2,\ldots,6n\}\;;$$ this means that $A_n$ contains the integers $k$ that satisfy the inequality $5n\le k\le 6n$. In particular,
$$\begin{align*}
A_1&=\{5,6\}\;,\\
A_2&=\{10,11,12\}\;,\\
A_3&=\{15,16,17,18\}\;,\\
A_4&=\{20,21,22,23,24\}\;,\text{ and}\\
A_5&=\{25,26,27,28,29,30\}\;.
\end{align*}$$
Thus,
$$A_1\cup A_2\cup A_3\cup A_4=\{5,6,10,11,12,15,16,17,18,20,21,22,23,24\}\;.$$
Now notice that $A_5$ picks up where $A_4$ leaves off: $\max A_4=24$, and $\min A_5=25$. And you can easily check that $A_6$ actually overlaps $A_5$: $\max A_5=30=\min A_6$.
In general we know from the definition that $\min A_n=5n$ and $\max A_n=6n$. Suppose that $n\ge 5$. Then $n-5\ge 0$, so
$$\max A_n=6n=5(n+1)+(n-5)\ge 5(n+1)=\min A_{n+1}\;.$$
In other words, if $n\ge 5$, the largest member of $A_n$ is always at least as large as the smallest member of $A_{n+1}$, so there’s no gap between $A_n$ and $A_{n+1}$, as there is between $A_1$ and $A_2$, between $A_2$ and $A_3$, and between $A_3$ and $A_4$. It follows that
$$\bigcup_{n\ge 5}A_n=\{25,26,\ldots\}=\{n\in\Bbb Z^+:n\ge 25\}$$
and hence that
$$\begin{align*}
\bigcup_{n\ge 1}A_n&=\{5,6,10,11,12,15,16,17,18,20,21,22,23,24\}\cup\{n\in\Bbb Z^+:n\ge 25\}\\
&=\{5,6,10,11,12,15,16,17,18\}\cup\{n\in\Bbb Z^+:n\ge 20\}\;.
\end{align*}$$
