Arithmetic progression within sequence: 1/2, 1/3, 1/4, 1/5, ......

Hint: The arithmetic progressions you found written in ascending order are:

$\dfrac{1}{6}, \dfrac{2}{6}, \dfrac{3}{6}$

$\dfrac{1}{12}, \dfrac{2}{12}, \dfrac{3}{12}, \dfrac{4}{12}$

$\dfrac{1}{60}, \dfrac{2}{60}, \dfrac{3}{60}, \dfrac{4}{60}, \dfrac{5}{60}$

See if you can generalize this. Spoiler below.

For any integer $n$, let $L_n = \text{lcm}(1,2,\ldots,n)$. Then, the numbers $\dfrac{1}{L_n}, \dfrac{2}{L_n}, \ldots, \dfrac{n}{L_n}$ are an arithmetic progression and are all in the sequence since $\dfrac{L_n}{n}$ is a positive integer. You could also use $\dfrac{1}{n!}, \dfrac{2}{n!}, \ldots, \dfrac{n}{n!}$.

Let $\frac1{a_1}, \frac1{a_2},\ldots ,\frac1{a_n}$ be a decreasing arithmetic progression of unit fractions of length $n$. Then you can add a rational number at the front, to make the arithmetic progression $$\frac k{a_0}, \frac1{a_1}, \frac1{a_2}, \ldots , \frac1{a_n}$$with $k, a_0$ being coprime integers. Now divide all of them by $k$, and you have a new arithmetic progression of unit fractions of length $n+1$. This you can keep going indefinitely.

You are, of course, free to divide by any multiple of $k$, but apart from that everything is forced, and given two fractions to start, whatever result you get will be similar (in the geometric sense) to what you get by dividing with exactly $k$. So essentially, the sequences you end up with are entirely decided by what two fractions you start with.