Question

100 баллов даю, помогите ​

Answer 1

$b_{n} =b_{1} {q}^{n - 1} \\ \displaystyle\bf\\\left \{ {{b_{3} +b_{6} = 1764 } \atop {b_{4} - b_{5} + b_{6} = 1323}} \right. \\ \displaystyle\bf\\\left \{ {{b_{1}q {}^{2} + b_{1} {q}^{5} = 1764 } \atop {b_{1}q {}^{3} - b_{1} {q}^{4} + b_{1} {q}^{5} = 1323 }} \right. \\ \displaystyle\bf\\\left \{ {{b_{1} {q}^{2}(1 + {q}^{3}) = 1764 } \atop {b_{1} {q}^{3} (1 - q + {q}^{2} ) = 1323}} \right. \\\displaystyle\bf\\ \div \left \{ {{b_{1} {q}^{2}(1 + q)(1 - q + {q}^{2} ) = 1764} \atop { b_{1} {q}^{3}(1 - q + {q}^{2} ) = 1323}} \right. \\ \\ \frac{b_{1} {q}^{2} (1 + q)(1 - q + {q}^{2}) }{b_{1} {q}^{3} (1 - q + {q}^{2}) } = \frac{1764}{1323} \\ \frac{1 + q}{q} = \frac{4}{3} \\ 4q = 3(1 + q) \\ 4q = 3 + 3q \\ 4q - 3q = 3 \\ q = 3 \\ \\ b_{1} \times {3}^{2} +b_{1} \times {3}^{5} = 1764 \\ 9b_{1} + 243b_{1} = 1764 \\ 252b_{1} = 1764 \\ b_{1} = 1764 \div 252 \\ b_{1} = 7 \\ \displaystyle\bf\\otvet \: \: \: \left \{ {{b_{1} = 7} \atop {q = 3 }} \right.$