已知公差大于零的等差数列{an}的前n项和为sn,且满足a3a4=117,a2+a5=22.求数列{an}的通项公式?
人气：418 ℃ 时间：2019-09-21 06:39:25
解答
等差数列中a2+a5=a3+a4,所以
a3a4=117,a3+a4=22,又因为公差大于0,解得a3=9,a4=13,公差d=4
所以an=a3+(n-3)*d=9+(n-3)*4=4n-3
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