3. Given an infinitely decreasing geometric progression. The sum of all its terms with odd indices is 10 more than the sum of all terms with even indices. And the difference between the sum of the squares of all terms at odd positions and the sum of the squares of all terms at even positions is 20. Find the first term and the common ratio of the progression.
Answer: $b_{1}=5, q=-\frac{1}{2}$.

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 $\frac{b_{1}}{1-q}$.

All terms of the given progression $\left\{b_{n}\right\}$ with odd indices also form a geometric progression with the first term $b_{1}$ and common ratio $q^{2}$, and all terms with even indices form a geometric progression with the first term $b_{2}=b_{1} q$ and common ratio $q^{2}$. The sums of these terms are $\frac{b_{1}}{1-q^{2}}$ and $\frac{b_{1} q}{1-q^{2}}$, respectively.

The squares of the terms of the progression at odd and even positions also form geometric progressions with the first term $b_{1}^{2}$ and common ratio $q^{2}$, and with the first term $b_{2}^{2}=b_{1}^{2} q^{2}$ and common ratio $q^{4}$, respectively. Their sums are $\frac{b_{1}^{2}}{1-q^{4}}$ and $\frac{b_{1}^{2} q^{2}}{1-q^{4}}$. From the condition, we obtain the system of equations

$$
\left\{\begin{array} { l } 
{ \frac { b _ { 1 } } { 1 - q ^ { 2 } } - \frac { b _ { 1 } q } { 1 - q ^ { 2 } } = 1 0 , } \\
{ \frac { b _ { 1 } ^ { 2 } } { 1 - q ^ { 4 } } - \frac { b _ { 1 } ^ { 2 } q ^ { 2 } } { 1 - q ^ { 4 } } = 2 0 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ \frac { b _ { 1 } ( 1 - q ) } { ( 1 - q ) ( 1 + q ) } = 1 0 , } \\
{ \frac { b _ { 1 } ^ { 2 } ( 1 - q ^ { 2 } ) } { ( 1 - q ^ { 2 } ) ( 1 + q ^ { 2 } ) } = 2 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\frac{b_{1}}{1+q}=10 \\
\frac{b_{1}^{2}}{1+q^{2}}=20
\end{array}\right.\right.\right.
$$

We square both sides of the first equation and then divide the second equation by the first term-by-term. We get $\frac{1+2 q+q^{2}}{1+q^{2}}=\frac{1}{5} \Leftrightarrow 4 q^{2}+10 q+4=0$, 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 valid. Then $b_{1}=10(1+q)=5$.