设等比数列{an}中,a1=256,前n项和为Sn,且Sn,Sn+2,Sn+1成等差数列,

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+2=Sn+Sn+1
a1[1-q^(n+1)]/(1-q)+a1[1-q^(n+1)]/(1-q)=2a1[1-q^(n+2)]/(1-q)
2q^2-q-1=0
(2q+1)(q-1)=0
q=-1/2 或q=1
T8=a1a2a3……a8=a1^8q^23
T9=a1a2a3……a9=a1^9q^31
T7=a1a2a3……a7=a1^q^16
当q=1时,T9>T8>T7
当q=-1/2时,T9