Question

在等比数列{an}中,a1+a2+a3=-3,a1*a2*a3=8,1、求an 2、求a1*a3*a5*a7*a9？

Answer 1

a1+a2+a3=-3,a1*a2*a3=8
a2/q+a2+a2*q=-3
a2^3=8,a2=2
2/q+2+2q=-3
2q+5+2/q=0
2q^2+5q+2=0
q=-1/2或-2,a1=-4或-1
an=-4*(-1/2)^(n-1)或an=-(-2)^(n-1)
a5=-1/4或-16
a1*a3*a5*a7*a9
=a5^5
=-1/2^10或-2^20,5,a1*a2*a3=8因为a2*a2=a1*a3,所以a2=2,a1=2/q,a3=2q,代入a1+a2+a3=-3得2/q+2+2q=-3
解得q=-2或q=-1/2
当q=-2时候a1=2/q=-1,an=a1*q^n=-(-2)^n,a2=2,a3=-4.a5=-16,a9=-64,a1*a3*a5*a7*a9=4096
当q=-1/2时候a1=2/q=-4,an=a1*q^n=-4*(-1/2)^n,a2=2,a3=-1,
a5=-1/16,a9=-1/64,a1*a3*a5*a7*a9=1/256,2,a1*a2*a3=a1*a1*q*a1*q^2=a1^3*q^3=8
a1*q=2
a1+a2+a3=a2/q+a2+a2*q=2/q+2+2q=-3
q=-2或-8
检验的-8不满足a1*a3*a5*a7*a9=a1*q^20=-1048576,0,