2. Determine the geometric progression of real numbers in which the sum of its first four terms is 15, and the sum of their squares is 85.
Solution. From the conditions of the problem, it follows that

$$
a\left(1+q+q^{2}+q^{3}\right)=15 \text { and } a^{2}\left(1+q^{2}+q^{4}+q^{6}\right)=85
$$

From this, it follows that

$$
\frac{\left(1+q+q^{2}+q^{3}\right)^{2}}{1+q^{2}+q^{4}+q^{6}}=\frac{45}{17} \text {, i.e., } \frac{q^{4}+2 q^{3}+2 q^{2}+2 q+1}{q^{4}+1}=\frac{45}{17} \text {. }
$$

The last equation is equivalent to the equation

$$
14 q^{4}-17 q^{3}-17 q^{2}-17 q+14=0
$$

However, $q \neq 0$, so we can divide the last equation by $q^{2} \neq 0$, after which we get the equation

$$
14\left(q^{2}+\frac{1}{q^{2}}\right)-17\left(q+\frac{1}{q}\right)-17=0
$$

which is equivalent to the equation

$$
14\left(q+\frac{1}{q}\right)^{2}-17\left(q+\frac{1}{q}\right)-45=0
$$

We introduce the substitution $t=q+\frac{1}{q}$ and get the equation $14 t^{2}-17 t-45=0$, whose solutions are $t_{1}=\frac{5}{2}$ and $t_{2}=-\frac{9}{7}$. For $t_{1}=\frac{5}{2}$, we get the equation $q+\frac{1}{q}=\frac{5}{2}$ whose solutions are $q_{1}=2, q_{2}=\frac{1}{2}$, while for $t_{2}=-\frac{9}{7}$, the equation $q+\frac{1}{q}=-\frac{9}{7}$ has no real solutions. Finally, for $q_{1}=2$, we find $a_{1}=1$, i.e., we get the progression $1,2,4,8, \ldots$, and for $q_{2}=\frac{1}{2}$, we find $a_{2}=8$, i.e., we get the progression $8,4,2,1, \frac{1}{2}, \ldots$.