Решебник по алгебре 9 класс Макарычев Задание 617

\[\boxed{\text{617\ (617).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[x_{1} = 3,\ \ x_{2} = 5,\ \ d = x_{2} - x_{1} = 2:\ \]

\[S_{n} = \frac{2x_{1} + d(n - 1)}{2} \cdot n = \frac{6 + 2 \cdot (n - 1)}{2} \cdot n = (3 + n - 1) \cdot n =\]

\[= (n + 2) \cdot n = n^{2} + 2n \leq 120;\]

\[n^{2} + 2n - 120 \leq 0\]

\[D = 1 + 120 = 121\]

\[так\ как\ n > 0 \Longrightarrow n = - 1 + 11 = 10.\]

\[Ответ:10\ членов.\]

\[\boxed{\text{617.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

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

\[S_{9} = c_{1} \cdot \frac{q^{9} - 1}{q - 1} = - 4 \cdot \frac{3^{9} - 1}{3 - 1} =\]

\[= - 2 \cdot 19682 = - 39\ 364;\]

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

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

\[= \frac{513}{3} = 171.\]

\[\textbf{в)}\ c_{1} = - 2;\ \ q = 2:\]

\[S_{9} = c_{1} \cdot \frac{q^{9} - 1}{q - 1} =\]

\[= - 2 \cdot \frac{(2)^{9} - 1}{2 - 1} =\]

\[= - 2 \cdot (512 - 1) = - 2 \cdot 511 =\]

\[= - 1022.\]

\[\textbf{г)}\ c_{1} = 32;\ \ q = - 0,5 = - \frac{1}{2}:\]

\[S_{9} = 32 \cdot \frac{\left( - \frac{1}{2} \right)^{9} - 1}{- \frac{1}{2} - 1} =\]

\[= 32 \cdot \frac{- \frac{1}{512} - 1}{- \frac{3}{2}} = \frac{32 \cdot 2 \cdot 513}{3 \cdot 512} =\]

\[= \frac{171}{8} = 21\frac{3}{8} = 21,375.\]