6.13 Find the natural numbers forming an arithmetic progression if the products of the first three and the first four of its terms are 6 and 24, respectively.
6.13 From the condition we have

$$
\begin{aligned}
& \left\{\begin{array}{l}
a_{1}\left(a_{1}+d\right)\left(a_{1}+2 d\right)=6 \\
a_{1}\left(a_{1}+d\right)\left(a_{1}+2 d\right)\left(a_{1}+3 d\right)=24
\end{array} \Rightarrow\right. \\
& \Rightarrow a_{1}+3 d=4 \Rightarrow a_{1}=4-3 d
\end{aligned}
$$

Thus, we arrive at the cubic equation

$3 d^{3}-22 d^{2}+48 d-29=0$, or

$3 d^{3}-3 d^{2}-19 d^{2}+19 d+29 d-29=0$,

from which $(d-1)\left(3 d^{2}-19 d+29\right)=0$. This equation has a single root $d=1$.

Answer: $1 ; 2 ; 3 ; 4$