Question

等比数列,a3=4,S3=12,求an通项公式

Answer 1

s3=a1+a2+a3

12=a1+a2+4

a1+a2=8

a1+a1q=8

a1(1+q)=8

a1=8/(1+q)

a1q^2=4

a1=4/q^2

8/(1+q)=4/q^2

2/(1+q)=1/q^2

2q^2=1+q

2q^2-q-1=0

(2q+1)(q-1)=0

q=-1/2或q=1

当q=-1/2时

a3=a1q^2

4=a1*1/4

a1=16

an=a1q^(n-1)

=16*(-1/2)^(n-1)

=(-1/2)^-4*(-1/2)^(n-1)

=(-1/2)^(n-5)

当q=1时

a3=a1q^2

4=a1*1

a1=4

an=a1q^(n-1)

=4*1^(n-1)

=4

所以当q=-1/2时,an=(-1/2)^(n-1)

当q=1时,an=4