设数列{an}的前n项和为sn,sn=a1(3^n-2)/2(n≥1),a4=54,则a1=
问题描述：
设数列{an}的前n项和为sn,sn=a1(3^n-2)/2(n≥1),a4=54,则a1=
答
等比数列前n项和公式Tn=a1(q^n-1)/(q-1)
观察Sn=a1(3^n-2)/2=a1(3^n-1)/(3-1)-0.5
即Sn为数列{an}的前n项和-0.5
所以Sn为等比数列,公比为3,所以a1=a4/(q^3)=54/27=2