25 , a1+a4=35 и a2+a3=30 прогрессия, найдите несколько членов прогрессии

A1+a4=35, a2+a3=30, q=?
a1+a1q³=35, a1q+a1q²=30
a1(1+q³)=35, a1(q+q²)=30
a1=35/(1+q³), a1=30/(q+q²)
35/(1+q³)=30/(q+q²)
35/30=(1+q³)/(q+q²)
7/6=(1+q)(1-q+q²)/q(q+1)
7/6=(1-q+q²)/q
7q=6(1-q+q²)
7q=6-6q+6q²
6q²-13q+6=0, D=169-144=25, √D=√25=5
q1=(13+5)/12=18/12=3/2
q2=(13-5)/12=8/12=2/3
a)q=3/2, a1=35/(1+27/8)=35/(35/8)=8
   a1=8,a2=12,a3=18,a4=27,
b)q=2/3, a1=35/(1+8/27)=35/(35/27)=27
   a1=27, a2=18, a3=12, a4=8,