已知数列{an}的前n项和为Sn,a1=1,若n≥2时,an是Sn与Sn-1的等差中项,则a5等于(  )A.18B.54C.16

2025-04-15 14:31:54
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回答1:

由题意知,2an=Sn+Sn-1(n≥2),
则2an+1=Sn+1+Sn
两式作差得:2an+1-2an=an+1+an,即an+1=3an(n≥2).
由a1=1,得a2=2.
an=2?3n?2(n≥2)
a5=2?33=54
故选:B.