4.035. When the thirteenth term of an arithmetic progression is divided by the third term, the quotient is 3, and when the eighteenth term is divided by the seventh term, the quotient is 2 and the remainder is 8. Determine the common difference and the first term of the progression.
Solution.

From the condition we have $\left\{\begin{array}{l}a_{13}=3 a_{3}, \\ a_{18}=2 a_{7}+8 .\end{array}\right.$

Using formula (4.1), we get

$$
\left\{\begin{array} { l } 
{ a _ { 1 } + 1 2 d = 3 ( a _ { 1 } + 2 d ) } \\
{ a _ { 1 } + 1 7 d = 2 ( a _ { 1 } + 6 d ) + 8 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ a _ { 1 } = 3 d } \\
{ a _ { 1 } = 5 d - 8 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
a_{1}=12 \\
d=4
\end{array}\right.\right.\right.
$$

Answer: $4,12$.

## Solutions to Chapter 6

## ALGEBRAIC EQUATIONS

## BASIC FORMULAS

For any $a, b$ and $c$, the following equalities are true:

$$
\begin{gathered}
(a+b)^{2}=a^{2}+2 a b+b^{2} \\
(a-b)^{2}=a^{2}-2 a b+b^{2} \\
a^{2}-b^{2}=(a-b)(a+b) \\
(a+b)^{3}=a^{3}+3 a^{2} b+3 a b^{2}+b^{3} \\
(a-b)^{3}=a^{3}-3 a^{2} b+3 a b^{2}-b^{3} \\
a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right) \\
a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)
\end{gathered}
$$

$a x^{2}+b x+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$, where $x_{1}, x_{2}$ are the roots of the equation

$$
a x^{2}+b x+c=0 \text {. }
$$

## SOLUTION OF EQUATIONS WITH ONE UNKNOWN

An equation with one unknown is an equality

$$
f_{