设数列{an}的前n项和Sn=a1(3n−1)2(n≥1,n∈N*)且a4=54,则a1=_.

问题描述:

设数列{an}的前n项和Sn=

a1(3n−1)
2
(n≥1,n∈N*)且a4=54,则a1=______.

由题意可得a4=S4-S3
=

a1(34−1)
2
-
a1(33−1)
2

=40a1-13a1=27a1=54,
解之可得a1=2
故答案为:2