The product of the third and the sixth terms of an arithmetic sequence is $406$. The ninth term of the sequence divided by the fourth term gives a quotient of $2$ and a remainder of $6$. Find the first term and the common difference of the arithmetic sequence.

Let $\{a_n\}$ be the arithmetic sequence so that we can construct an equation for each of the statements above.

Let $a_3$ be the third term of the sequence and $a_6$ be the sixth term of the sequence.
Then their product is 
$a_3.a_6=406$...………....………… (1)
And
Let $a_9$ be the ninth term of the sequence, use the formula, Dividend=Divisor.Quotient+Remainder 
$a_9=2.a_4+6$ …....…...…………....(2)
Rewrite (1) as
$(a_1+2d)(a_1+5d)=406$........…..…....……..(3)
And rewrite (2) as 
$a_1+8d=(a_1+3d)2+6$…....…………......... (4)
Simplify (4)
$a_1+8d=2a_1+6d+6$
$a_1=2d-6$….…....……....…….(5)
Substitute (5) into (3) to get
$(4d-6)(7d-6)=406$
$14d^2-33d-185=0$
Solve the equation using quadratic formula 
$d=5$ and $d=\frac{-37}{28}$
Two different values for a common difference will give us two different arithmetic sequences:
For $d=5$ in (5)
$a_1=4$
And for $d=\frac{-37}{28}$ in (5)
$a_1=\frac{-79}{7}$
QED. 

References 

Ellina Grigoriva (2016), Methods of Solving Sequence and Series Problems, Springer international publishing AG. Page 15