已知等差数列5,427,347…,记第n项到第n+6项的和为Tn,则|Tn|取得最小值时,n的值为()A.5B.6C.7D.8-高二数学

题目简介

已知等差数列5,427,347…,记第n项到第n+6项的和为Tn,则|Tn|取得最小值时,n的值为()A.5B.6C.7D.8-高二数学

题目详情

已知等差数列5,4
2
7
,3
4
7
…,记第n项到第n+6项的和为Tn,则|Tn|取得最小值时,n的值为(  )
A.5B.6C.7D.8
题型:单选题难度:中档来源:不详

答案

首项a1=5,公差d=-class="stub"5
7

an=5+(n-1)(-class="stub"5
7
)
=-class="stub"5
7
n+class="stub"40
7
an+6=-class="stub"5
7
(n+6)+class="stub"40
7
=-class="stub"5
7
n+class="stub"10
7

所以Tn=
[(-class="stub"5
7
n+class="stub"40
7
)+(-class="stub"5
7
n+class="stub"10
7
)]×7
2
=-5n+25,
所以当n=5时,|Tn|取得最小值0,
故选A.

更多内容推荐