Question

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

Answer 1

b1+b4=b1+b1q³=b1(1+q³)=36⇒b1=36/(1+q³)
b1+b2=b1+b1q=b1(1+q)=18⇒b1=18/(1+q)
36/(1+q)(1-q+q²)=18/(1+q)
1-q+q²=2
q²-q-1=0
D=1+4=5
q1=(1-√5)/2⇒b1=18:(1+(1-√5)/2)=36/(3-√5)⇒b5=36/(3-√5)*(1-√5)^4/16=
=36/(3-√5)*4(3-√5)²/16=9(3-√5)
q1=(1+√5)/2⇒b1=18:(1+(1+√5)/2)=36/(3+√5)⇒b5=9(3+√5

Answer 2

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

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