Найти пятый член геометрической прогресс,в которой b1+b4=36.b1+b2=18.

Решите задачу:

$\left \{ {{b_1+b_4=36} \atop {b_1+b_2=18}} \right. \; \left \{ {{b_1+b_1q^3=36} \atop {b_1+b_1q=18}} \right. \; \left \{ {{b_1(1+q^3)=36} \atop {b_1(1+q)=18}} \right. \\\\\frac{36}{1+q^3}=\frac{18}{1+q}\\\\36(1+q)=18(1+q)(1^2-q+q^2)\, |:18(1+q)\ne 0\\\\1-q+q^2=2\\\\q^2-q-1=0\\\\D=1+4=5\\\\q_1=\frac{1-\sqrt5}{2},\; q_2=\frac{1+\sqrt5}{2}\\\\(b_1)_1=\frac{18}{1+\frac{1+\sqrt5}{2}}=\frac{36}{3-\sqrt5}$

$(b_1)_2=\frac{18}{1+\frac{1+\sqrt5}{2}}=\frac{36}{3+\sqrt5}\\\\(b_5)_1=b_1\cdot q^4=\frac{36}{3-\sqrt5}\cdot (\frac{1-\sqrt5}{2})^4\\\\(b_5)_2=\frac{36}{3+\sqrt5}\cdot (\frac{1+\sqrt5}{2})^4$