Question

геометрическое прогрессияa1+a2+a3=19 и a1^2+a2^2+a3^2=133 найдите a1, a2, a3, a4

Answer 1

$a_{1}+a_{1}q+a_{1}q^2=19\\ a_{1}^2+a_{1}^2q^2+a_{1}^2q^4=133\\\\ \frac{a_{1}^2(1+q^2+q^4)}{a_{1}(1+q+q^2)}=\frac{133}{19}\\ a_{1}\frac{1+q^2+q^4}{1+q+q^2}=7\\ a_{1}(q^2-q+1)=7\\ \frac{19}{1+q+q^2}=\frac{7}{q^2-q+1}\\ 19q^2-19q+19=7+7q+7q^2\\ 12q^2-26q+12=0\\ 6q^2-13q+6=0\\ D=5^2\\ q=\frac{13+5}{12}=\frac{3}{2}\\ q=\frac{13-5}{12}=\frac{2}{3}\\ a_{1}=4\\ a_{2}=6\\ a_{3}=9\\\\ a_{1}=9\\ a_{2}=6\\ a_{3}=4$