3. It is known that the ratio of the sum of all terms of an infinitely decreasing geometric progression to the sum of the cubes of all terms of the same progression is $\frac{1}{12}$, and the ratio of the sum of the fourth powers of all terms to the sum of the squares of all terms of this progression is $\frac{36}{5}$. Find the first term and the common ratio of the specified progression.
Answer: $b_{1}= \pm 3, q=-\frac{1}{2}$.

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

$$
\left\{\begin{array} { l } 
{ \frac { b _ { 1 } ^ { 3 } } { 1 - q ^ { 3 } } : \frac { b _ { 1 } } { 1 - q } = 1 2 } \\
{ \frac { b _ { 1 } ^ { 4 } } { 1 - q ^ { 4 } } : \frac { b _ { 1 } ^ { 2 } } { 1 - q ^ { 2 } } = \frac { 3 6 } { 5 } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\frac{b_{1}^{2}}{1+q+q^{2}}=12 \\
\frac{b_{1}^{2}}{1+q^{2}}=\frac{36}{5}
\end{array}\right.\right.
$$

Dividing the first equation by the second term by term, we get $\frac{1+q^{2}}{1+q+q^{2}}=\frac{5}{3} \Leftrightarrow 2 q^{2}+5 q+2=0$, from which $q=-2$ or $q=-\frac{1}{2}$. Since the progression is decreasing and infinite, $|q|<1$, and only the value $q=-\frac{1}{2}$ is suitable. Then $b_{1}^{2}=\frac{36}{5}\left(1+q^{2}\right)=9$ and $b_{1}= \pm 3$.