已知等比数列{an}中,a1+a2+a3=18,a2+a3+a4=-9,求a1+a2+a3+a4+a5

Question

如题所述

Answer 1

a1(1+q+q^2)=18
a1*q(1+q+q^2)=-9
两式相除,得：q=-1/2
所以a1=24,a2=-12,a3=6,a4=-3,a5=3/2
所以a1+a2+a3+a4+a5
=24-12+6-3+3/2
=33/2