Прогрессия a1,a1q,a1q²,...
a1+a1q=1 1/3⇒a1(1+q)=4/3⇒a1=4/3(1+q)
[(a1)³+(a1q)³+(a1q²)³+....]/[(a1)2+(a1q)2+(a1q²)2+....]=12/13
1)S1=[(a1)³+(a1q)³+(a1q²)³+....]
b1=(a1)³ q1=q³
S1=b1/(1-q1)=(a1)³/(1-q³)
2)S2=[(a1)²+(a1q)²+(a1q²)²+....]
c1=(a1)³ q2=q³
S2=c1/(1-q2)=(a1)²/(1-q²)
(a1)³/(1-q³):(a1)²/(1-q²)=12/13
(a1)³/(1-q)(1+q+q²)*(1-q)(1+q)/(a1)²=12/13
a1*(1+q)/(1+q+q²)=12/13
a1=12(1+q+q²)/13(1+q)
12(1+q+q²)/13(1+q)=4/3(1+q)
3(1+q+q²)/13=1/3
9(1+q+q²)=13
9+9q+9q²-13=0
9q²+9q-4=0
D=81+144=225
a)q=(-9-15)/18=-4/3
a1=4:[3(1-4/3)]=4:(-1)=-4
а3=а1*q²=-4*16/9=-64/9
b)q=(-9+15)/18=1/3
a1=4:[3(1+1/3)]=4:4=1
a3=1*1/9=1/9
Ответ a3=-64/9;a3=1/9