設(shè)
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是等差數(shù)列
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的前
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項和,若
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,則數(shù)列
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的通項為
分析:由題意可得a
4=a
1+3d=9,s
3=3a
1+
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d=15,求出a
1=3,d=2,代入等差數(shù)列的通項公式可求答案.
解答:解:由題意知a
4=a
1+3d=9,s
3=3a
1+
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d=15,
兩者聯(lián)立解得a
1=3,d=2,
a
n=3+(n-1)×2=2n+1,
故答案為 a
n=2n+1.
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