设递增等差数列{an}的前n项和为Sn,已知a3=1,a4是a3和a7的等比中项, (I)求数列{an}的通项公式; (II)求数列{an}的前n项和Sn.
问题描述:
设递增等差数列{an}的前n项和为Sn,已知a3=1,a4是a3和a7的等比中项,
(I)求数列{an}的通项公式;
(II)求数列{an}的前n项和Sn.
答
(Ⅰ)设等差数列{an}的首项为a1,公差为d(d>0),由a3=1得,a1+2d=1①,由a4是a3和a7的等比中项得,(a1+3d)2=(a1+2d)(a1+6d)②,整理②得,2a1d+3d2=0,因为d>0,所以2a1+3d=0③,联立①③得:a1=-3,d=2.所...