Question

设等比数列an的公比为q,前n项和为sn,若s(n+1),sn,s(n+2)成等差数列,求q的值

Answer 1

若q=1,则S(n+1)=n+1,Sn=n,S(n+2)=n+2,

此时S(n+1),Sn,S(n+2)不成等差数列

所以q≠1,则Sn=a1*(1-q^n)/(1-q)

a1*[1-q^(n+1)]/(1-q)+a1*[1-q^(n+2)]/(1-q)=2*a1*(1-q^n)/(1-q)

[1-q^(n+1)]+[1-q^(n+2)]=2-2*q^n

q^(n+1)+q^(n+2)=2*q^n

q+q²=2

q²+q-2=0

(q+2)(q-1)=0

∴q=-2