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,则q的值为

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,则q的值为

Answer 1

当公比是1时,是常数列,an=a1
sn=n*a1,sn+1=(n+1)*a1,sn+2=(n+2)*a1
若S(n+1),Sn,S(n+2)成等差
2Sn=S(n+1)+S(n+2)
2n*a1=(n+1)*a1+(n+2)*a1
a1=0
不成立
当公比不是一时Sn=a1(1-q^n)/1-q
Sn+1=a1(1-q^n+1)/1-q
Sn+2=a1(1-q^n+2)/1-q
2Sn=S(n+1)+S(n+2)
2a1(1-q^n)/1-q==a1(1-q^n+1)/1-q +a1(1-q^n+2)/1-q
2a1(1-q^n)=a1(1-q^n+1)+a1(1-q^n+2)
2-2q^n)=2-q^n+1-q^n+2
2q^n=q^n+1+q^n+2
解方程即可
2=q+q2
q=-2或q=1（舍）