Задача 72640 Пом.пож алгебра....
Условие
Решение
[m]\left\{\begin {matrix}b_{2}+b_{4}=45\\b_{2}\cdot b_{4}=324\end {matrix}\right.[/m]
[m]b_{n}=b_{1}\cdot q^{n-1}[/m]
[m]b_{2}=b_{1}\cdot q^{2-1}[/m]
[m]b_{4}=b_{1}\cdot q^{4-1}[/m]
[m]\left\{\begin {matrix}b_{1}\cdot q+b_{1}\cdot q^3=45\\b_{1}\cdot q\cdot b_{1}\cdot q^3=324\end {matrix}\right.[/m] ⇒ [m]\left\{\begin {matrix}b_{1}\cdot q(1+ q^2)=45\\b^2_{1}\cdot q^4=324\end {matrix}\right.[/m] ⇒ [m]\left\{\begin {matrix}b_{1}\cdot q(1+ q^2)=45\\b_{1}\cdot q^2=18\end {matrix}\right.[/m] ⇒ [m]\left\{\begin {matrix}b_{1}\cdot q=\frac{45}{1+q^2}\\b_{1}\cdot q=\frac{18}{q}\end {matrix}\right.[/m]
[m]\frac{45}{1+q^2}=\frac{18}{q}[/m] ⇒ [m]\frac{5}{1+q^2}=\frac{2}{q}[/m] ⇔[m] 2(1+q^2)=5q[/m] - квадратное уравнение
[m]2q^2-5q+2=0[/m]
D=25-4*2*2=25-16=9
[m]q=2[/m] или [m] q=\frac{1}{2}[/m]( не удовл условию)
[m]b_{1}=\frac{18}{q^2}[/m]
[m]b_{1}=\frac{18}{2^2}=4,5[/m]
21.
[m]a_{3}=2a_{1}[/m]
[m]S_{5}=190[/m]
[m]a_{n}=a_{1}+d\cdot(n-1)[/m]
[m]a_{3}=a_{1}+d\cdot(3-1)[/m] ⇒ [m]a_{3}=a_{1}+2d[/m]
[m]a_{1}+2d=2a_{1}[/m]
[m]S_{n}=\frac{2a_{1}+d(n-1)}{2}\cdot n[/m]
[m]S_{5}=\frac{2a_{1}+d(5-1)}{2}\cdot 5[/m]
[m]S_{5}=190[/m]
[m]\frac{2a_{1}+d(5-1)}{2}\cdot 5=190[/m]
[m]\left\{\begin {matrix}a_{1}+2d=2a_{1}\\\frac{2a_{1}+d(5-1)}{2}\cdot 5=190\end {matrix}\right.[/m] ⇒ [m]\left\{\begin {matrix}a_{1}=2d\\\frac{2\cdot 2d+d(5-1)}{2}\cdot 5=190\end {matrix}\right.[/m] ⇒ [m]\left\{\begin {matrix}a_{1}=2d\\4d+4d=76\end {matrix}\right.[/m]
[m]d=9,5[/m]
22.
[m]b_{4}=8b_{1}[/m]
[m]b_{3}+b_{4}+14=b_{3}\cdot b_{4}[/m]
[m]\left\{\begin {matrix}b_{4}=8b_{1}\\b_{3}+b_{4}+14=b_{3}\cdot b_{4}\end {matrix}\right.[/m]
[m]b_{n}=b_{1}\cdot q^{n-1}[/m]
[m]b_{3}=b_{1}\cdot q^{3-1}[/m] ⇒ [m]b_{3}=b_{1}\cdot q^{2}[/m]
[m]b_{4}=b_{1}\cdot q^{4-1}[/m] ⇒ [m]b_{4}=b_{1}\cdot q^{3}[/m]
[m]\left\{\begin {matrix}b_{1}\cdot q^{3}=8b_{1}\\b_{1}\cdot q^{2}+b_{1}\cdot q^{3}+14=b_{1}\cdot q^{2}\cdot b_{1}\cdot q^{3}\end {matrix}\right.[/m] ⇒ [m]\left\{\begin {matrix} q^{3}=8 ⇒q=2 \\b_{1}\cdot 2^{2}+b_{1}\cdot 2^{3}+14=b_{1}\cdot 2^{2}\cdot b_{1}\cdot 2^{3}\end {matrix}\right.[/m] ⇒ [m]\left\{\begin {matrix}q=2 \\32b^2_{1}-12b_{1}-14=0\end {matrix}\right.[/m]
D=144-4*32*(-14)=144+1792=1936=44^2
[m]b_{1}=\frac{56}{64}=\frac{7}{8}=0,875[/m] или [m]b_{1}=-\frac{32}{64}[/m] ( не явл. положительным)