若等差數(shù)列{an}中,公差d=2,且a1+a2+…+a100=200,則a5+a10+a15+…+a100的值是 .
【答案】
分析:由a
1+a
2+…+a
100=200可知,s
100=
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=
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=200,求出a
1得到a
n的通項,a
5+a
10+a
15+…+a
100為等差數(shù)列,求出首項和公差即可求出之和.
解答:解:因為s
100=
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=
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=200,所以a
1=-97,則a
n=2n-99,
而a
5,a
10,a
15,…,a
100為首項為-89,公差為10的等差數(shù)列,項數(shù)是20,
則a
5+a
10+a
15+…+a
100=20×(-89)+
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=120
故答案為120
點評:考查學(xué)生靈活運用等差數(shù)列的前n項和的公式,靈活運用等差數(shù)列性質(zhì)的能力.