b1(1+q^3)=54
b1q(1+q)=36
36(1+q^2-q)=54q
2q^2-5q+2=0
q=1/2
b1=54/(1+1/8)=48
s=b1/(1-q)=48/(1-1/2)=96
b1(1+q^3)=54
b1q(1+q)=36
36(1+q^2-q)=54q
2q^2-5q+2=0
q=1/2
b1=54/(1+1/8)=48
s=b1/(1-q)=48/(1-1/2)=96
b1+b1q^3=54⇒b1(1+q^3)=54⇒b1(1+q)(1-q+q^2)=54
b1q+b1q^2=36⇒b1q(1+q)=36
Получаем систему:
b1(1+q)(1-q+q^2)=54
b1q(1+q)=36⇒b1=36/(q^2+q)
Делим первое уравнение на второе
(1-q+q^2)/q=54/36⇒(1-q+q^2)/q=3/2⇒
2(1-q+q^2)=3q⇒2q^2-5q+2=0⇒
D=5^2-4*2*2=25-16=9; √D=3
q1=(5+3)/4=2; q2=(5-3)/4=1/2
Так как прогрессия бесконечно убывающая, то
q=1/2
b1=36/(q^2+q)=36/(1/4+1/2)=36/(3/4)=36*4/3=48
b1=48; b2=48*1/2=24; b3=b2*q=24*1/2=12; b4=b3*q=12*1/2=6