若等差数列{an}中,公差d=2,且a1+a2+…+a100=200,则a5+a10+a15+…+a100的值是 .
【答案】
分析:由a
1+a
2+…+a
100=200可知,s
100=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131023214355220566116/SYS201310232143552205661010_DA/0.png)
=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131023214355220566116/SYS201310232143552205661010_DA/1.png)
=200,求出a
1得到a
n的通项,a
5+a
10+a
15+…+a
100为等差数列,求出首项和公差即可求出之和.
解答:解:因为s
100=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131023214355220566116/SYS201310232143552205661010_DA/2.png)
=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131023214355220566116/SYS201310232143552205661010_DA/3.png)
=200,所以a
1=-97,则a
n=2n-99,
而a
5,a
10,a
15,…,a
100为首项为-89,公差为10的等差数列,项数是20,
则a
5+a
10+a
15+…+a
100=20×(-89)+
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131023214355220566116/SYS201310232143552205661010_DA/4.png)
=120
故答案为120
点评:考查学生灵活运用等差数列的前n项和的公式,灵活运用等差数列性质的能力.