{an}是等比数列,a1+a2+a3=3,s4+s5+s6=6,求an

1个回答

  • 设公比为q.

    q=1时,sn=n,s4+s5+s6=4+5+6=156,不符.

    q1时:

    s3=a1(1-q^3)/(1-q)=3

    s4+s5+s6=a1(1-q^4+1-q^5+1-q^6)/(1-q)=a1(3-q^4-q^5-q^6)/(1-q)=6

    两式相除得:(3-q^4-q^5-q^6)/(1-q^3)=2

    即:

    q^6+q^5+q^4-2q^3-1=0

    q^6-q^5+2q^5-2q^4+3q^4-3q^3+q^3-1=0

    (q-1)(q^5+2q^4+3q^3+q^2+q+1)=0

    (q^5+2q^4+3q^3+q^2+q+1)=0

    解得唯一实根q=-0.728676677160428

    a1=3(1-q)/(1-q^3)=3.739282176466450