Question

在等比数列中a3=3/2,s3=9/2，求q和a1

Answer 1

S3/a3=(a1+a2+a3)/a3
=(a3/q²+a3/q +a3)/a3
=1/q²+1/q+1
=(9/2)/(3/2)
=3
2q²-q-1=0
(q-1)(2q+1)=0
q=1或q=-1/2
q=1时，a1=a3/q²=(3/2)/1=3/2
q=-1/2时，a1=a3/q²=(3/2)/(-1/2)²=6

a1=3/2 q=1或a1=6 q=-1/2

追问

能把每一步都给我说清楚吗 我看你的答案有很多地方都不懂

Answer 2

s3=a1+a2+a3
=a3/q²+a3/q+a3
=(3/2)*(1/q²+1/q+1)
=9/2
1/q²+1/q+1=3
(1/q-1)(1/q+2)=0
1/q=1或1/q=-2
即q=1或q=-1/2
(1)当q=1时
a1=a3/q²=3/2
(2)当q=-1/2时
a1=a3/q²=(3/2)/(1/4)=6
述上所述
q=1、a1=3/2
或q=-1/2、a1=6