4.004. Find the first three terms $a_{1}, a_{2}, a_{3}$ of an arithmetic progression, given that $a_{1}+a_{3}+a_{5}=-12$ and $a_{1} a_{3} a_{5}=80$.
Solution.

From the condition we have $\left\{\begin{array}{l}a_{1}+a_{3}+a_{5}=-12, \\ a_{1} \cdot a_{3} \cdot a_{5}=80 .\end{array}\right.$

Using formula (4.1), we get

$$
\begin{aligned}
& \left\{\begin{array}{l}
a_{1}+a_{1}+2 d+a_{1}+4 d=-12, \\
a_{1}\left(a_{1}+2 d\right)\left(a_{1}+4 d\right)=80
\end{array} \Leftrightarrow\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
a_{1}+2 d=-4, \\
a_{1}\left(a_{1}+2 d\right)\left(a_{1}+2 d+2 d\right)=80
\end{array} \Leftrightarrow\right.
\end{aligned}
$$

$$
\begin{aligned}
& \Leftrightarrow\left\{\begin{array} { l } 
{ a _ { 1 } = - 4 - 2 d , } \\
{ a _ { 1 } \cdot ( - 4 ) ( - 4 + 2 d ) = 8 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
a_{1}=-4-2 d, \\
a_{1}(d-2)=-10
\end{array} \Leftrightarrow\right.\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ a _ { 1 } = - 4 - 2 d , } \\
{ ( - 4 - 2 d ) ( d - 2 ) = - 1 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
a_{1}=-4-2 d, \\
d^{2}=9
\end{array} \Leftrightarrow\right.\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
a_{1}=-4-2 d, \\
d= \pm 3 .
\end{array}\right.
\end{aligned}
$$

From here $\left\{\begin{array}{l}a_{1}=-4-2 d, \\ d=-3\end{array}\right.$ or $\left\{\begin{array}{l}a_{1}=-4-2 d, \\ d=3,\end{array}\right.$

T.e.

$$
\left\{\begin{array} { l } 
{ a _ { 1 } = 2 , } \\
{ d = - 3 }
\end{array} \text { or } \left\{\begin{array}{l}
a_{1}=-10, \\
d=3 .
\end{array}\right.\right.
$$

Then

$$
a_{1}^{\prime}=2, a_{2}^{\prime}=2-3=-1, a_{3}^{\prime}=2-6=-4 ;
$$

or

$$
a_{1}^{\prime \prime}=-10, a_{2}^{\prime \prime}=-10+3=-7, a_{3}^{\prime \prime}=-10+6=-4
$$

Answer: 1) $2,-1,-4 ; 2)-10,-7,-4$.