[m] b_{n}=b_{1}q^{n-1}[/m]
[m] b_{1}+b_{1}q+b_{1}q^2+....+b_{1}q^{n-1}+...=56[/m]
[m] b_{1}(1+q+q^2+....+q^{n-1}+...)=56[/m]⇒
Так как прогрессия убывающая, то ее сумма [m]S=b_{1}\frac{1}{1-q}[/m]
[m]b_{1}\cdot \frac{1}{1-q}=56[/m]
Аналогично составляем вторую сумму:
[m] b^2_{1}+b^2_{1}q^2+b^2_{1}(q^2)^2+....=448[/m]⇒ [m]b^2_{1}\cdot \frac{1}{1-q^2}=448[/m]
Решаем систему:
[m]\left\{\begin {matrix}b_{1}\cdot \frac{1}{1-q}=56\\b^2_{1}\cdot \frac{1}{1-q^2}=448\end {matrix}\right.[/m] ⇒ [m]\left\{\begin {matrix}b_{1}= \frac{56}{\frac{1}{1-q}}\\(\frac{56}{\frac{1}{1-q}})^2\cdot \frac{1}{1-q^2}=448\end {matrix}\right.[/m] ⇒ [m]\left\{\begin {matrix}b_{1}= 56(1-q)\\( 56(1-q))^2\cdot \frac{1}{1-q^2}=448\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}b_{1}= 56(1-q)\\3136(1-q)^2\cdot \frac{1}{(1-q)(1+q)}=448\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}b_{1}= 56(1-q)\\7(1-q)\cdot \frac{1}{(1+q)}=1\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}b_{1}= 56(1-q)\\7-7q=1+q\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}b_{1}= 56(1-q)\\6=8q\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}b_{1}= 56(1-\frac{3}{4})\\q=\frac{3}{4}\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}b_{1}= 14\\q=\frac{3}{4}\end {matrix}\right.[/m]
О т в е т. D)