已知等差数列{an}中,a2=7,a4=15,则前10项的和S10=( ) A.100  B.210

已知等差数列{an}中,a2=7a4=15,则前10项的和S10=(  )

A100  B210  C380  D400

答案

B考点】等差数列的通项公式.

【分析】由第二项和第四项的值可以求出首项和公差,写出等差数列前n项和公式,代入n=10得出结果.

【解答】解:d=a1=3

∴S10=10×3+\frac{10×9×4}{2}

=210

故选B

【点评】若已知等差数列的两项,则等差数列的所有量都可以求出,只要简单数字运算时不出错,问题可解.

 

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