Question

在等比数列an中，已知a1＋ a2＋ a3＝ 6,a2 ＋a3＋ a4＝ － 3,则a3 ＋a4 ＋a5 ＋a6＋ a7＋ a8＝

要详细过程

Answer 1

设数列公比为q
a2 ＋a3＋ a4＝ q(a1＋ a2＋ a3)
q=(a2 ＋a3＋ a4)/(a1＋ a2＋ a3)=-3/6=-1/2
a3 ＋a4 ＋a5 ＋a6＋ a7＋ a8
=q(a2 ＋a3＋ a4)+q^4(a2 ＋a3＋ a4)
=(a2 ＋a3＋ a4)(q+q^4)
=(-3)*[(-1/2+(-1/2)^4]
=21/16

Answer 2

a1+a2+a3=a1(1+q+q^2)=6 (1)
a2+a3+a4=a1q(1+q+q^2)=-3 (2)
(2)/(1)
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)+a2q^4(1+q+q^2)
=6q^2-3q^4
=6/4-3/8
=9/8

Answer 3

-3=a2 ＋a3＋ a4=q(a1＋ a2＋ a3)＝ 6
所以 公比 q=-1/2
a4 ＋a5＋ a6=q^3(a1＋ a2＋ a3)=-1/8*6=-3/4
a3 ＋a4 ＋a5 ＋a6＋ a7＋ a8=q^2(a1＋ a2＋ a3+a4 ＋a5＋ a6)=21/16