4.026. Find the natural numbers forming an arithmetic progression if the product of the first three and the first four of its terms are 6 and 24, respectively.
Solution.

From the condition we have $\left\{\begin{array}{l}a_{1}\left(a_{1}+d\right)\left(a_{2}+d\right)=6, \\ a_{1}\left(a_{1}+d\right)\left(a_{1}+2 d\right)\left(a_{1}+3 d\right)=24\end{array} \Rightarrow\right.$

$$
\Rightarrow \frac{a_{1}\left(a_{1}+d\right)\left(a_{2}+d\right)}{a_{1}\left(a_{1}+d\right)\left(a_{2}+d\right)\left(a_{1}+3 d\right)}=\frac{6}{24} \Rightarrow a_{1}+3 d=4, a_{1}=4-3 d
$$

We obtain the equation

$$
\begin{aligned}
& 3 d^{3}-22 d^{2}+48 d-29=0 \Leftrightarrow 3 d^{3}-3 d^{2}-19 d^{2}+19 d+29 d-29=0 \Leftrightarrow \\
& \Leftrightarrow(d-1)\left(3 d^{2}-19 d+29\right)=0 \Rightarrow d_{1}=1, d_{2,3}=\frac{19 \pm \sqrt{13}}{6} \text { (do not fit). } \\
& d=1 \Rightarrow a_{1}=1, a_{2}=2, a_{3}=3, a_{4}=4
\end{aligned}
$$

Answer: $1,2,3,4, \ldots$