题目

已知正项数列{an},其前n项和Sn满足10Sn=an2+5an+6，且a1，a3，a15成等比数列，求数列{an}的通项an. 答案：解:∵10Sn=an2+5an+6,                         ① ∴10a1=a12+5a1+6,解之得a1=2或a1=3.又10Sn-1=an-12+5an-1+6(n≥2),           ②由①-②得10an=(an2-an-12)+6(an-an-1),即(an+an-1)(an-an-1-5)=0.∵an+an-1＞0,∴an-an-1=5(n≥2).当a1=3时,a3=13,a15=73.a1,a3,a15不成等比数列,∴a1≠3.当a1=2时,a3=12,a15=72,有a32=a1a15.∴a1=2.∴an=5n-3.

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