4.022. The product of the first three terms of a geometric progression is 1728, and their sum is 63. Find the first term and the common ratio of this progression.
## Solution.

From the condition, we have $\left\{\begin{array}{l}b_{1} \cdot b_{2} \cdot b_{3}=1728, \\ b_{1}+b_{2}+b_{3}=63 .\end{array}\right.$

Using formula (4.6), we get
$\left\{\begin{array}{l}b_{1} \cdot b_{1} q \cdot b_{1} q^{2}=1728, \\ b_{1}+b_{1} q+b_{1} q^{2}=63\end{array} \Leftrightarrow\left\{\begin{array}{l}\left(b_{1} q\right)^{3}=1728, \\ b_{1}+b_{1} q+\left(b_{1} q\right) q=63\end{array} \Leftrightarrow\left\{\begin{array}{l}b_{1} q=12, \\ b_{1}+12+12 q=63\end{array} \Leftrightarrow\right.\right.\right.$ $\Leftrightarrow\left\{\begin{array}{l}b_{1} q=12, \\ b_{1}+12 q=51 .\end{array} \Rightarrow 4 q^{2}-17 q+4=0\right.$

$q=4$ or $q=\frac{1}{4}$.

From this, we get $\left\{\begin{array}{l}q=4, \\ b_{1}=3\end{array}\right.$ or $\left\{\begin{array}{l}q=\frac{1}{4}, \\ b_{1}=48 .\end{array}\right.$

Answer: 1) 3, 4; 2) 48, $\frac{1}{4}$.