6.14 The product of the third and sixth terms of an arithmetic progression is 406. When the ninth term of this 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.
6.14 According to the condition, we have the system

$$
\left\{\begin{array} { l } 
{ a _ { 3 } a _ { 6 } = 4 0 6 , } \\
{ a _ { 9 } = 2 a _ { 4 } + 6 , }
\end{array} \text { or } \left\{\begin{array}{l}
\left(a_{1}+2 d\right)\left(a_{1}+5 d\right)=406 \\
a_{1}+8 d=2\left(a_{1}+3 d\right)+6
\end{array}\right.\right.
$$

From here, we find two solutions: 1) $a_{1}=4, d=5$;
2) $a_{1}=-\frac{79}{7}, d=-\frac{37}{14}$. Only the first solution satisfies the condition of the problem.

Answer: 4 and 5.