Решебник по алгебре 9 класс Макарычев Задание 577

\[\boxed{\text{577}\text{\ (577)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ c_{1} = 20;\ \ d = 3:\]

\[c_{n} = 20 + 3 \cdot (n - 1);\ \]

\[c_{5} = 20 + 3 \cdot 4 = 20 + 12 = 32.\]

\[\textbf{б)}\ c_{1} = 5,8;\ \ \ d = - 1,5:\]

\[c_{n} = 5,8 - 1,5 \cdot (n - 1);\]

\[c_{21} = 5,8 - 1,5 \cdot 20 = - 24,2.\ \]

\[\boxed{\text{577.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[x_{1} = 21,\ \ d = - 0,5:\ \text{\ \ \ \ \ \ \ \ \ \ }\]

\[x_{6} = 21 - 0,5 \cdot (6 - 1) = 18,5;\]

\[x_{25} = 21 - 0,5 \cdot (25 - 1) =\]

\[= 21 - 12 = 9;\ \]

\[S_{n} = \frac{\left( x_{1} + x_{n} \right)}{2} \cdot n;\]

\[S_{20} = \frac{18,5 + 9}{2} \cdot 20 =\]

\[= 27,5 \cdot 10 = 275.\]