4.016. Find four numbers that form a geometric progression, where the second term is 35 less than the first, and the third term is 560 more than the fourth.
## Solution.

From the condition, we have $\left\{\begin{array}{l}b_{1}-b_{2}=35, \\ b_{3}-b_{4}=560 .\end{array}\right.$

Using formula (4.6), we get

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ b _ { 1 } - b _ { 1 } q = 3 5 , } \\
{ b _ { 1 } q ^ { 2 } - b _ { 1 } q ^ { 3 } = 5 6 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
b_{1}(1-q)=35, \\
b_{1} q^{2}(1-q)=560
\end{array} \Rightarrow \frac{1-q}{q^{2}(1-q)}=\frac{35}{560} \Rightarrow\right.\right. \\
& \Rightarrow q^{2}=16, q_{1}=-4, q_{2}=4 . \text { Substituting } q_{1}=-4, \text { we get }
\end{aligned}
$$

$$
b_{1}^{\prime}=7, b_{2}^{\prime}=-28, b_{3}^{\prime}=112, b_{4}^{\prime}=-448 . \text { Substituting } q_{2}=4 \text {, we get }
$$

$b_{1}^{\prime \prime}=-\frac{35}{3}, b_{2}{ }^{\prime \prime}=-\frac{140}{3}, b_{3}{ }^{\prime \prime}=-\frac{560}{3}, b_{4}{ }^{\prime \prime}=-\frac{2240}{3}$.

Answer: 1) 7, - 28, 112, $-448 ; 2)-\frac{35}{3},-\frac{140}{3},-\frac{560}{3},-\frac{2240}{3}$.