3. Find the first term and the common ratio of an infinite decreasing geometric progression, if the ratio of the sum of the cubes of all its terms to the sum of all terms of this progression is $\frac{48}{7}$, and the ratio of the sum of the fourth powers of the terms to the sum of the squares of the terms of this progression is $\frac{144}{17}$.
Answer: $b_{1}= \pm 3, q=\frac{1}{4}$.

Solution. It is known that the sum of the first $n$ terms of a geometric progression with the first term $b_{1}$ and common ratio $q$ is $\frac{b_{1}\left(1-q^{n}\right)}{1-q}$. For a decreasing infinite geometric progression, $|q|<1$, so as $n$ approaches infinity, $q^{n}$ approaches zero, and the sum of the terms approaches $\frac{b_{1}}{1-q}$.

The cubes of the terms of the given progression $\left\{b_{n}\right\}$ also form a geometric progression with the first term $b_{1}^{3}$ and common ratio $q^{3}$, the fourth powers of the terms form a progression with the first term $b_{1}^{4}$ and common ratio $q^{4}$, and the squares form a progression with the first term $b_{1}^{2}$ and common ratio $q^{2}$. The sums of these terms are respectively $\frac{b_{1}^{3}}{1-q^{3}}, \frac{b_{1}^{4}}{1-q^{4}}$, and $\frac{b_{1}^{2}}{1-q^{2}}$.

From the condition, we obtain the system of equations

$$
\left\{\begin{array} { l } 
{ \frac { b _ { 1 } ^ { 3 } } { 1 - q ^ { 3 } } : \frac { b _ { 1 } } { 1 - q } = \frac { 4 8 } { 7 } } \\
{ \frac { b _ { 1 } ^ { 4 } } { 1 - q ^ { 4 } } : \frac { b _ { 1 } ^ { 2 } } { 1 - q ^ { 2 } } = \frac { 1 4 4 } { 1 7 } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\frac{b_{1}^{2}}{1+q+q^{2}}=\frac{48}{7} \\
\frac{b_{1}^{2}}{1+q^{2}}=\frac{144}{17}
\end{array}\right.\right.
$$

Dividing the first equation by the second term by term, we get $\frac{1+q^{2}}{1+q+q^{2}}=\frac{17}{21} \Leftrightarrow 4 q^{2}-17 q+4=0$, from which $q=4$ or $q=\frac{1}{4}$. Since the progression is decreasing and infinite, $|q|<1$, and only the value $q=\frac{1}{4}$ is suitable. Then $b_{1}^{2}=\frac{144}{17}\left(1+q^{2}\right)=9$ and $b_{1}= \pm 3$.