6.20 The sum of an infinite geometric progression with a common ratio $|q|<1$ is 4, and the sum of the cubes of its terms is 192. Find the first term and the common ratio of the progression.
6.20 Since

$$
\begin{aligned}
& b_{1}+b_{1} q+b_{1} q^{2}+\ldots=\frac{b_{1}}{1-q} \\
& b_{1}^{3}+b_{1}^{3} q^{3}+b_{1}^{3} q^{6}+\ldots=\frac{b_{1}^{3}}{1-q^{3}}
\end{aligned}
$$

$$
\left\{\begin{array} { l } 
{ \frac { b _ { 1 } } { 1 - q } = 4 , } \\
{ \frac { b _ { 1 } ^ { 3 } } { 1 - q ^ { 3 } } = 192 , }
\end{array} \text { or } \left\{\begin{array}{l}
4(1-q)=b_{1} \\
48\left(1+q+q^{2}\right)=b_{1}^{2}
\end{array}\right.\right.
$$

Further, we have $2 q^{2}+5 q+2=0$, from which $q=-2$ (not suitable), $q=-0.5$. Therefore, $b_{1}=6$.

Answer: 6 and $-0.5$.