4.025. 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.
## Solution.

From the condition, we have $\left\{\begin{array}{l}b_{1}+b_{2}+b_{3}+\ldots=16, \\ b_{1}^{2}+b_{2}^{2}+b_{3}^{2}+\ldots=153.6,\end{array}\right.$

Using formula (4.6), we get

$$
\left\{\begin{array} { l } 
{ b _ { 1 } + b _ { 1 } q + b _ { 1 } q ^ { 2 } + \ldots = 1 6 , } \\
{ b _ { 1 } ^ { 2 } + b _ { 1 } ^ { 2 } q ^ { 2 } + b _ { 1 } ^ { 2 } q ^ { 4 } + \ldots = 1 5 3 . 6 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
b_{1}\left(1+q+q^{2}+\ldots\right)=16 \\
b_{1}^{2}\left(1+q^{2}+q^{4}+\ldots\right)=153.6
\end{array}\right.\right.
$$

Using formula (4.12), we get

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

$\Rightarrow b_{1}=16(1-q) \Rightarrow(16(1-q))^{2} \cdot \frac{1}{1-q^{2}}=153.6$, from which $q=\frac{1}{4}$. Then $b_{1}=16\left(1-\frac{1}{4}\right)=12$. Using formula (4.6), we get $b_{4}=b_{1} q^{3}=12 \cdot \frac{1}{64}=\frac{3}{16}$. Answer: $\frac{3}{16}, \frac{1}{4}$.