4.019. Find the first term and the common ratio of the geometric progression, given that $b_{4}-b_{2}=-45 / 32$ and $b_{6}-b_{4}=-45 / 512$.
## Solution.

From the condition, we have $\left\{\begin{array}{l}b_{4}-b_{2}=-\frac{45}{32}, \\ b_{6}-b_{4}=-\frac{45}{512} .\end{array}\right.$

Using formula (4.6), we get

$$
\left\{\begin{array} { l } 
{ b _ { 1 } q ^ { 3 } - b _ { 1 } q = - \frac { 4 5 } { 3 2 } , } \\
{ b _ { 1 } q ^ { 5 } - b _ { 1 } q ^ { 3 } = - \frac { 4 5 } { 5 1 2 } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
b_{1} q\left(q^{2}-1\right)=-\frac{45}{32}, \\
b_{1} q^{3}\left(q^{2}-1\right)=-\frac{45}{512}
\end{array} \Rightarrow \frac{q\left(q^{2}-1\right)}{q^{3}\left(q^{2}-1\right)}=\frac{512}{32} \Rightarrow\right.\right.
$$

$\Rightarrow q^{2}=\frac{1}{16}, q_{1}=-\frac{1}{4}, q_{2}=\frac{1}{4}$. Substituting these values of $q_{1}$ and $q_{2}$, we find $b_{1}^{\prime}=-6$ and $b_{1}^{\prime \prime}=6$.

Answer: 1) $\left.-6,-\frac{1}{4} ; 2\right) 6, \frac{1}{4}$.