Question

已知等比数列{An}的公比为q,前n项的和为Sn,且S3,S9,S6成等差数列.

q知道后，试问A4,A7的等差中项是数列{An}中的第几项

Answer 1

由已知,可得
S3=A1(1-q^3)/(1-q);
S9=A1(1-q^9)/(1-q);
S6=A1(1-q^6)/(1-q);
S3,S9,S6成等差数列,所以
S3+S6=2S9,化简,得
q^3+q^6=2q^9
q^3(2q^3+1)(q^3-1)=0,
解得,q^3=0(舍去),或q^3=1,或q^3=-1/2
当q^3=1时,An=A1,A4,A7的等差中项是数列{An}中的任一项.
当q^3=-1/2,即q=-2^(-1/3)时,
A4=A1*q^3=-A1/2,
A7=A1*q^6=A1/4,
所以,A4,A7的等差中项为:(A4+A7)/2=-A1/8,
An=A1*q^(n-1)=-A1*2^[-(n-1)/3]=-A1*2^(-1/3),
解得,
n=2