Question

等比数列〔an〕中，已知a1+a2+a3=6,a2+a3+a4=一3，则a3+a4+…+a8=？

急求答案

Answer 1

a1+a2+a3=a1(1+q+q^2)=6
a2+a3+a4=a2(1+q+q^2)=-3

a2/a1=(-3)/6=-1/2

q=-1/2

a3+a4+a5+a6+a7+a8
=a3(1+q+q^2)+a6(1+q+q^2)
=a1q^2(1+q+q^2)+a1q^5(1+q+q^2)
=a1(1+q+q^2)(q^2+q^5)
=6*[(-1/2)^2+(-1/2)^5]
=6*(1/4-1/32)
=21/16