ГДЗ по алгебре 9 класс Макарычев Задание 690

\[\boxed{\text{690}\text{\ (690)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ a_{1} = 2,\ \ d = 2,\]

\[a_{n} = a_{1} + d(n - 1) = 2 + 2n - 2 = 2n\]

\[2n = 200 \Longrightarrow n = 100;\]

\[S_{100} = \frac{a_{1} + a_{100}}{2} \cdot 100 = (2 + 200) \cdot 50 = 10100.\]

\[\textbf{б)}\ a_{1} = 1,\ \ d = 2,\ \ \]

\[a_{n} = a_{1} + d(n - 1) = 1 + 2n - 2 = 2n - 1\]

\[2n - 1 = 149\]

\[2n = 150 \Longrightarrow n = 75,\]

\[S_{75} = \frac{a_{1} + a_{75}}{2} \cdot 75 = \frac{1 + 149}{2} \cdot 75 = 5625.\]

\[\textbf{в)}\ a_{1} = 102,\ \ d = 3,\]

\[a_{n} = a_{1} + d(n - 1) = 102 + 3n - 3 = 3n + 99 = 198\]

\[3n + 99 = 198\]

\[3n = 99\]

\[n = 33,\]

\[S_{33} = \frac{a_{1} + a_{33}}{2} \cdot 33 = \frac{102 + 198}{2} \cdot 33 = 150 \cdot 33 = 4950.\]