ГДЗ по алгебре 9 класс Макарычев Задание 684

\[\boxed{\text{684}\text{\ (684)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ a_{1} = \frac{2}{3};\ \ \ a_{2} = \frac{3}{4} \Longrightarrow d = a_{2} - a_{1} = \frac{3}{4} - \frac{2}{3} = \frac{1}{12};\]

\[S_{10} = \frac{2a_{1} + d(n - 1)}{2} \cdot n = \frac{2 \cdot \frac{2}{3} + \frac{1}{12} \cdot (10 - 1)}{2} \cdot 10 = \left( \frac{4}{3} + \frac{3}{4} \right) \cdot 5 =\]

\[= \frac{16 + 9}{12} \cdot 5 = \frac{25}{12} \cdot 5 = \frac{125}{12} = 10\frac{5}{12}.\]

\[\textbf{б)}\ a_{1} = \sqrt{3},\ \ a_{2} = \sqrt{12},\ \]

\[\Longrightarrow d = a_{2} - a_{1} = \sqrt{12} - \sqrt{3} = \sqrt{3} \cdot (2 - 1) = \sqrt{3};\]

\[S_{10} = \frac{2a_{1} + d(n - 1)}{2} \cdot n = \left( 2\sqrt{3} + 9\sqrt{3} \right) \cdot 5 = 11\sqrt{3} \cdot 5 = 55\sqrt{3}.\]