ГДЗ по алгебре и начала математического анализа 11 класс Алимов Задание 1447

\[\boxed{\mathbf{1447}\mathbf{.}}\]

\[b(n) - геометрическая\ \]

\[прогрессия;\]

\[q - знаменатель.\]

\[1)\ b_{3} - b_{2} = 9\]

\[b_{1}q^{2} - b_{1} = 9\]

\[b_{1} \bullet \left( q^{2} - 1 \right) = 9\]

\[b_{1} = \frac{9}{q^{2} - 1}.\]

\[2)\ b_{2} - b_{4} = 18\]

\[b_{1}q - b_{1}q^{3} = 18\]

\[b_{1} \bullet \left( q - q^{3} \right) = 18\]

\[b_{1} = \frac{18}{q - q^{3}}.\]

\[Составим\ уравнение:\]

\[\frac{9}{q^{2} - 1} = \frac{18}{q - q^{3}};\ \ q \neq \pm 1\]

\[9\left( q - q^{3} \right) = 18\left( q^{2} - 1 \right)\]

\[q - q^{3} = 2q^{2} - 2\]

\[q^{3} + 2q^{2} - q - 2 = 0\]

\[q^{2} \bullet (q + 2) - (q + 2) = 0\]

\[(q + 2)\left( q^{2} - 1 \right) = 0\]

\[q_{1} = - 2;\ \ \ q_{2} = \pm 1.\]

\[q = - 2:\]

\[b_{1} = \frac{9}{( - 2)^{2} - 1} = \frac{9}{4 - 1} = \frac{9}{3} = 3;\]

\[b_{2} = 3 \bullet ( - 2) = - 6;\]

\[b_{3} = - 6 \bullet ( - 2) = 12;\]

\[b_{4} = 12 \bullet ( - 2) = - 24.\]

\[Ответ:\ \ 3;\ \ - 6;\ \ 12;\ \ - 24.\]