Question

高一数学数列

Answer 1

a[3]=3/2
S[3]=9/2
S[2]=9/2-3/2=3=a[1]+a[2]=3/2×(1/q)^2+3/2×(1/q)
3q^2-3/2 q-3/2 =0
6q^2-3q-3=0
(2q+1)(q-1)=0
q=1或-1/2
a[1]=3/2或6

追答

行家正解，满意请速采

追问

谢谢，一直没看到

Answer 2

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

(q-1)(2q+1)=0
q=1或q=-½
q=1时，a1=a3=3/2
q=-½时，a1=a3/q²=(3/2)/(-½)²=6
a1的值为3/2或6

Answer 3

a3=a1q^2=3/2,
S3=a1+a2+a3=9/2
a1+a2=3
a1(1+q)=3.............(1)
a1q^2=3/2...........(2)
两式相除得
2q^2-q-1=0
q=-1/2或q=1
a1=6或a1=3/2