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МатематикаГДЗ по алгебре 9 класс Макарычев Задание 654
Задание 654
\[\boxed{\text{654\ (654).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ x_{5} = x_{1}q^{4} = 1\frac{1}{9},\ \ q = \frac{1}{3}:\ \ \]
\[\frac{10}{9} = x_{1} \cdot \frac{1}{81};\ \ \ \ \ \ x_{1} = 90\ \ \ \ \]
\[S_{5} = x_{1} \cdot \frac{q^{5} - 1}{q - 1} = 90 \cdot \frac{\frac{1}{243} - 1\ }{\frac{1}{3} - 1\ } = \frac{90 \cdot 242 \cdot 3}{2 \cdot 243} = 134\frac{4}{9}.\]
\[\textbf{б)}\ x_{4} = x_{1} \cdot q^{3} = 121,5,\ \ q = - 3:\ \ \]
\[121,5 = x_{1} \cdot ( - 3)^{3};\ \ \ \ \ x_{1} = - 4,5\]
\[S_{5} = - 4,5 \cdot \frac{( - 3)^{5} - 1}{- 3 - 1} = - \frac{244 \cdot 9}{4 \cdot 2} = - 274,5.\]
\[Ответ:а)\ 134\frac{4}{9};\ \ б) - 274,5.\]
\[\boxed{\text{654.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[\textbf{а)}\ d = - 0,4,\ \ n = 12,\ \ \]
\[a_{n} = 2,4 \Longrightarrow\]
\[a_{n} = a_{1} + d \cdot (n - 1)\]
\[2,4 = a_{1} - 0,4 \cdot (12 - 1)\]
\[2,4 = a_{1} - 4,4\]
\[a_{1} = 6,8\]
\[S_{12} = \frac{a_{1} + a_{12}}{2} \cdot 12 =\]
\[= (6,8 + 2,4) \cdot 6 = 55,2.\]
\[\textbf{б)}\ a_{1} = - 35,\ \ d = 5,\]
\[\ \ S = 250 \Longrightarrow\]
\[S_{n} = \frac{2a_{1} + d(n - 1)}{2} \cdot n =\]
\[= \frac{- 70 + 5n - 5}{2} \cdot n = 250\]
\[- 75n + 5n^{2} = 100\]
\[n^{2} - 15n - 100 = 0\]
\[D = 15^{2} + 4 \cdot 100 = 625\]
\[n = \frac{15 \pm 25}{2} = 20;\ - 5;\]
\[так\ как\ n > 0,\ то\ n = 20;\]
\[a_{20} = a_{1} + d(n - 1) =\]
\[= - 35 + 5 \cdot 19 = 60.\]
\[\textbf{в)}\ d = \frac{1}{2},\ \ a_{n} = 50,\]
\[\text{\ \ }S_{n} = 2525 \Longrightarrow\]
\[S_{n} = \frac{a_{1} + 50}{2} \cdot n = 2525\]
\[n\left( a_{1} + 50 \right) = 5050\]
\[\left\{ \begin{matrix} a_{1} + 0,5 \cdot (n - 1) = 50 \\ n\left( a_{1} + 50 \right) = 5050\ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} a_{1} = 50,5 - 0,5n\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ n(50,5 - 0,5n + 50) = 5050 \\ \end{matrix} \right.\ \]
\[\Longrightarrow 0,5n^{2} - 100,5n + 5050 =\]
\[= 0 \Longrightarrow n^{2} - 201n + 10100 =\]
\[= 0 \Longrightarrow\]
\[D = 201^{2} - 4 \cdot 10100 = 1\]
\[n = \frac{201 \pm 1}{2},\ \ n_{1} = 100,\ \ \]
\[n_{2} = 101;\]
\[1)\ если\ \ n = 100,\]
\[\text{\ \ }a_{1} = a_{n} - d(n - 1) =\]
\[= 50 - \frac{1}{2} \cdot 99 = 0,5;\]
\[2)\ если\ n = 101,\ \ \]
\[a_{1} = a_{n} - d(n - 1) =\]
\[= 50 - \frac{1}{2} \cdot 100 = 0;\]
\[\textbf{г)}\ a_{1} = - \frac{1}{2},\]
\[\text{\ \ }a_{n} = a_{1} + d(n - 1) =\]
\[= - 0,5 + d(n - 1) = - 29\frac{1}{2};\]
\[S_{n} = \frac{a_{1} + a_{n}}{2} \cdot n =\]
\[= \frac{- \frac{1}{2} - 29\frac{1}{2}}{2} \cdot n = - 450;\]
\[\Longrightarrow - \frac{30}{2} \cdot n = - 450,\ \ n = 30,\]
\[\Longrightarrow - 0,5 + 29d = - 29,5\]
\[29d = - 29\]
\[d = - 1.\]
