Question

在等比数列中,已知a1+a2+a3=6,a2+a3+a4=-3,则a3+a4+a5+a6+a7=?

Answer 1

设公比为q

a1+a2+a3=6

a2+a3+a4=q(a1+a2+a3)=-3

所以q=-1/2

a1+a2+a3

=a1+a1*q+a1*q^2

=a1(1+q+q^2)

=3*a1/4=6

a1=8

a3=a1*q^2=2

a3+a4+a5+a6+a7

=a3*(1+q)+q^4(a1+a2+a3)

=1+3/8

=11/8