Решебник по алгебре 9 класс Макарычев Задание 623

 

\[\boxed{\text{623}\text{\ (623)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ b_{1} = 6;\ \ q = 2:\ \ \]

\[b_{2} = b_{1} \cdot q = 6 \cdot 2 = 12,\ \ \]

\[b_{3} = b_{2} \cdot q = 12 \cdot 2 = 24,\ \ \]

\[b_{4} = b_{3} \cdot q = 24 \cdot 2 = 48,\ \ \]

\[b_{5} = b_{4} \cdot q = 48 \cdot 2 = 96.\]

\[\textbf{б)}\ b_{1} = - 16;\ \ q = \frac{1}{2}:\]

\[b_{2} = b_{1} \cdot q = - 16 \cdot \frac{1}{2} = - 8,\ \ \]

\[b_{3} = b_{2} \cdot q = - 8 \cdot \frac{1}{2} = - 4,\ \ \]

\[b_{4} = b_{3} \cdot q = - 4 \cdot \frac{1}{2} = - 2,\ \ \]

\[b_{5} = b_{4} \cdot q = - 2 \cdot \frac{1}{2} = - 1\text{.\ }\]

\[\textbf{в)}\ \ b_{1} = - 24;\ \ q = - 1,5:\ \]

\[b_{2} = b_{1} \cdot q = - 24 \cdot ( - 1,5) = 36,\ \ \]

\[b_{3} = b_{2} \cdot q = 36 \cdot ( - 1,5) = - 54,\ \]

\[b_{4} = b_{3} \cdot q = - 54 \cdot ( - 1,5) = 81,\ \ \]

\[b_{5} = b_{4} \cdot q = 81 \cdot ( - 1,5) = - 121,5.\]

\[\textbf{г)}\ \ b_{1} = 0,4;\ \ \ q = \sqrt{2}:\ \]

\[b_{2} = b_{1} \cdot q = 0,4 \cdot \sqrt{2},\ \ \]

\[b_{3} = b_{2} \cdot q = 0,4 \cdot \sqrt{2} \cdot \sqrt{2} = 0,8,\ \ \]

\[b_{4} = b_{3} \cdot q = 0,8\sqrt{2},\ \ \]

\[b_{5} = b_{4} \cdot q = 0,8 \cdot \sqrt{2} \cdot \sqrt{2} = 1,6.\]

\[\boxed{\text{623.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[b_{2} = 6,\ \ b_{4} = 54,\ \ q > 0;\]

\[b_{4} = b_{2} \cdot q^{2}\]

\[q^{2} = \frac{b_{4}}{b_{2}}\]

\[q^{2} = \frac{54}{6} = 9\]

\[q = 3\]

\[q = - 3 \Longrightarrow не\ подходит\ \]

\[по\ условию.\]

\[\Longrightarrow b_{2} = b_{1} \cdot q \Longrightarrow b_{1} = \frac{b_{2}}{q} \Longrightarrow\]

\[\Longrightarrow b_{1} = \frac{6}{3} = 2.\]

\[S_{7} = \frac{b_{1} \cdot \left( q^{7} - 1 \right)}{q - 1} =\]

\[= \frac{2 \cdot (3^{7} - 1)}{3 - 1} = 2186.\]