ГДЗ по алгебре 9 класс Макарычев Задание 677

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\[\textbf{а)}\ a_{1} = 9\sqrt{3} - 2,\ \ d = 2 - \sqrt{3},\]

\[a_{12} = a_{1} + d(12 - 1) = a_{1} + 11d = 9\sqrt{3} - 2 + 22 - 11\sqrt{3} = 20 - 2\sqrt{3}.\]

\[\textbf{б)}\ a_{1} = \frac{5\sqrt{3} - 7}{3},\ \ d = \frac{\sqrt{3} - 2}{3},\]

\[a_{8} = a_{1} + d(8 - 1) = a_{1} + 7d = \frac{5\sqrt{3} - 7}{3} + 7 \cdot \frac{\sqrt{3} - 2}{3} =\]

\[= \frac{12\sqrt{3} - 21}{3} = 4\sqrt{3} - 7.\]