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