6.12 The sum of an infinite geometric progression with a common ratio $|q|<1$ is 16, and the sum of the squares of the terms of this progression is 153.6. Find the fourth term and the common ratio of the progression.
6.12 Applying formula (6.10) to both progressions, we get

$$
\begin{aligned}
& b_{1}+b_{1} q+b_{1} q^{2}+\ldots=\frac{b_{1}}{1-q} \\
& b_{1}^{2}+\left(b_{1} q\right)^{2}+\left(b_{1} q^{2}\right)^{2}+\ldots=\frac{b_{1}^{2}}{1-q^{2}}
\end{aligned}
$$

According to the condition, we have the system

$\left\{\begin{array}{l}\frac{b_{1}}{1-q}=16, \\ \frac{b_{1}^{2}}{1-q^{2}}=153.6,\end{array}\right.$

from which $q=\frac{1}{4}, b_{1}=12$. Therefore, $b_{4}=b_{1} q^{3}=12 \cdot \frac{1}{64}=\frac{3}{16}$.

Answer: $\frac{3}{16}$ and $\frac{1}{4}$.