248. Can a geometric progression with $k$ terms be selected from the infinite sequence $1 ; \frac{1}{2} ; \frac{1}{3} ; \ldots ; \frac{1}{n} ; \ldots$ for any natural number $k>2$?

$3^{*}$

67
$\triangle$ This can be done, and in more than one way. For example, a geometric progression $1 ; \frac{1}{2} ; \frac{1}{4} ; \ldots ; \frac{1}{2^{k-1}}$ would work.

Answer: it is possible.