Найдите первый член и знаменатель геометрической прогрессии (bn), если: b3+b6=420 и b4-b5+b6=315.

Ответ:

\[\left\{ \begin{matrix} b_{3} + b_{6} = 420\ \ \ \ \ \ \ \ \ \\ b_{4} - b_{5} + b_{6} = 315 \\ \end{matrix}\text{\ \ \ \ \ \ \ } \right.\ \]

\[\left\{ \begin{matrix} b_{1}q^{2} + b_{1}q^{5} = 420\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ b_{1}q^{3} - q^{4}b_{1} + b_{1}q^{5} = 315 \\ \end{matrix}\text{\ \ \ \ \ \ \ } \right.\ \]

\[\left\{ \begin{matrix} b_{1}q^{2}\left( 1 + q^{3} \right) = 420\ \ \ \ \ \ \ \\ b_{1}q^{3}\left( 1 - q + q^{2} \right) = 315 \\ \end{matrix} \right.\ \]

\[\frac{1 + q}{q} = \frac{4}{3}\]

\[3 + 3q = 4q\]

\[q = 3.\]

\[b_{1} = \frac{420}{q^{2}\left( 1 + q^{3} \right)} = \frac{420}{9 \cdot 28} =\]

\[= \frac{15}{9} = \frac{5}{3} = 1\frac{2}{3}.\]

\[Ответ:\ \ q = 3;\ \ \ b_{1} = 1\frac{2}{3}.\]