设数列{an}的公差为d(d≠0),首项为a1,
由已知得:
(3a1+3d)2=9(2a1+d)
4a1+6d=4(2a1+d).
解之得:
a1=
4
9
d=
8
9或
a1=0
d=0(舍)
∴an=a1+(n−1)d=
4
9+(n−1)×
8
9=
4
9(2n−1).
设数列{an}的公差为d(d≠0),首项为a1,
由已知得:
(3a1+3d)2=9(2a1+d)
4a1+6d=4(2a1+d).
解之得:
a1=
4
9
d=
8
9或
a1=0
d=0(舍)
∴an=a1+(n−1)d=
4
9+(n−1)×
8
9=
4
9(2n−1).