4.032. The product of the third and sixth terms of an arithmetic progression is 406. When the ninth term of the progression is divided by its fourth term, the quotient is 2, and the remainder is 6. Find the first term and the common difference of the progression.
## Solution.

From the condition, we have $\left\{\begin{array}{l}a_{3} \cdot a_{6}=406, \\ a_{9}=2 a_{4}+6 .\end{array}\right.$ Using formula (4.1), we get

$$
\begin{aligned}
& \left\{\begin{array}{l}
\left(a_{1}+2 d\right)\left(a_{1}+5 d\right)=406, \\
a_{1}+8 d=2\left(a_{1}+3 d\right)+6
\end{array}\right. \\
& \Rightarrow a_{1}=2 d-6 \text { and } 14 d^{2}-33 d-185=0
\end{aligned}
$$

from which we find

$$
\begin{aligned}
& d_{1}=-\frac{37}{14}, d_{2}=5 \text {. Then } a_{1}^{\prime}=2\left(-\frac{37}{14}\right)-6=-\frac{79}{7} \text { (does not fit) or } \\
& a_{1}^{\prime \prime}=2 \cdot 5-6=4 . \\
& \text { Answer: } 4 \text { and } 5 .
\end{aligned}
$$