Решебник по алгебре 9 класс Мерзляк Задание 717

\[\boxed{\mathbf{717\ (717).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[a_{n} = a_{1} + d(n - 1)\]

\[a_{1} = 4;\ \ \ \ \ d = 0,4\]

\[1)a_{3} = 4 + 0,4 \cdot 2 =\]

\[= 4 + 0,8 = 4,8\]

\[2)\ a_{11} = 4 + 0,4 \cdot 10 =\]

\[= 4 + 4 = 8\]

\[3)\ a_{32} = 4 + 0,4 \cdot 31 =\]

\[= 4 + 12,4 = 16,4\]

\[\boxed{\mathbf{717.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[S_{n} = 6 \cdot \left( \left( - \frac{1}{2} \right)^{n} - 1 \right)\]

\[b_{1} = S_{1} = 6 \cdot \left( - \frac{1}{2} - 1 \right) =\]

\[= 6 \cdot \left( - \frac{3}{2} \right) = - 9\]

\[S_{2} = 6 \cdot \left( \left( - \frac{1}{2} \right)^{2} - 1 \right) =\]

\[= 6 \cdot \left( \frac{1}{4} - 1 \right) = 6 \cdot \left( - \frac{3}{4} \right) = \ - \frac{9}{2}\]

\[так\ как\ \ \ b_{1} + b_{2} = S_{2}\]

\[\ b_{2} = S_{2} - b_{1} = - \frac{9}{2} - ( - 9) =\]

\[= - \frac{9}{2} + 9 = \frac{9}{2}\]

\[q = \frac{9}{2}\ :( - 9) = - \frac{1}{2}\]

\[b_{4} = b_{1} \cdot q^{3} = - 9 \cdot \left( - \frac{1}{2} \right)^{3} =\]

\[= - 9 \cdot \left( - \frac{1}{8} \right) = \frac{9}{8}\]

\[Ответ:b_{4} = \frac{9}{8}.\]