Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 981

\[\left( b_{n} \right) - геометрическая\ \]

\[прогрессия;\ \]

\[\left( a_{n} \right) - арифметическая\ \]

\[прогрессия.\]

\[1)\ a_{1} = b_{5} = b_{1} \bullet q^{4};\]

\[a_{2} = b_{8} = b_{1} \bullet q^{7};\]

\[a_{10} = b_{11} = b_{1} \bullet q^{10}.\]

\[2)\ d = a_{2} - a_{1} = b_{1}q^{7} - b_{1}q^{4} =\]

\[= b_{1}q^{4} \bullet \left( q^{3} - 1 \right).\]

\[3)\ a_{10} = a_{1} + 9d =\]

\[= b_{1}q^{4} + 9b_{1}q^{4}\left( q^{3} - 1 \right);\]

\[b_{1}q^{10} = b_{1}q^{4}\left( 1 + 9q^{3} - 9 \right);\]

\[q^{6} = 9q^{3} - 8\]

\[q^{6} - 9q^{3} + 8 = 0\]

\[D = 81 - 32 = 49\]

\[q_{1}^{3} = \frac{9 - 7}{2} = 1;\]

\[q_{2}^{3} = \frac{9 + 7}{2} = 8;\]

\[q_{1} = \sqrt[3]{1} = 1;q_{2} = \sqrt[3]{8} = 2.\]

\[4)\ q = 1:\]

\[S_{5} = 5b_{1} = 62;\]

\[b_{1} = \frac{62}{5} = 12,4.\]

\[5)\ q = 2:\]

\[S_{5} = \frac{b_{1}\left( q^{5} - 1 \right)}{q - 1} = 62;\]

\[b_{1}\left( 2^{5} - 1 \right) = 62\]

\[b_{1} \bullet 31 = 62\]

\[b_{1} = 2.\]

\[Ответ:\ \ 12,4\ или\ 2.\]