Из условия задачи, имеем
b1+b1q=108 => b1(1+q)=108
b1q+b1q^2=135 => b1(q+q^2)=135
из первого уравнения получаем
b1=108/(1+q) , q не равно -1
Подставим во второе уравнение
(108/(1+q))*(q+q^2)=135
108(q+q^2)=135(1+g)
108q^2+108q-135q-135=0
108q^2-27q-135=0
4q^2-g-5=0
Решая это квадратное уравнение, получаем корни
q=-1 - не удовлетворяет ОДЗ
q=1,25
тогда b1=108/(1+q)=108/2,25=48
1 член прогрессии = b1=48
2- = b1q=48*1,25=60
3- =b1q^2=48*(1.25)^2=75
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