Решебник по алгебре 9 класс Макарычев Задание 650

\[\boxed{\text{650\ (}\text{с}\text{).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ c_{1} = - 4,\ \ q = 3,\ \ \]

\[S_{9} = c_{1} \cdot \frac{q^{9} - 1}{q - 1} = - 4 \cdot \frac{3^{9} - 1}{3 - 1} = - 2 \cdot 19682 = - 39\ 364.\]

\[\textbf{б)}\ c_{1} = 1,\ \ q = - 2,\]

\[S_{9} = c_{1} \cdot \frac{q^{9} - 1}{q - 1} = \frac{( - 2)^{9} - 1}{- 2 - 1} = \frac{513}{3} = 171.\]

\[Ответ:а) - 39364;\ \ б)\ 171.\]

\[\boxed{\text{650.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ a_{20} = 1,7,\ \ a_{37} = 0,\]

\[\ \ d = ?\]

\[a_{37} = a_{20} + 16d\]

\[0 = 1,7 + 17d\]

\[17d = - 1,7\]

\[d = - 0,1.\]

\[\textbf{б)}\ a_{10} = 270,\ \ d = - 3,\]

\[\text{\ \ }a_{100} = ?\]

\[a_{100} = a_{10} + 90d =\]

\[= 270 - 3 \cdot 90 = 0.\]