>> MRPA151373DDT2A100 pdf Tama Electric
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Tama Electric>> MRPA151432FDT1A100 pdf Tama Electric
MRPA151432FDT1A100 价格
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Tama Electric>> MRPA151960DDT2A100 pdf Tama Electric
MRPA151960DDT2A100 价格
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型号：MRPA151960DDT2A100
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Tama Electric若等差数列{an}中，公差d=2，且a1+a2+…+a100=200，则a5+a10+a15+…+a100的值是
迷离小胤子QY15
因为s100=1+a100)&2=1+198)2=200，所以a1=-97，则an=2n-99，而a5，a10，a15，…，a100为首项为-89，公差为10的等差数列，项数是20，则a5+a10+a15+…+a100=20×（-89）+=120故答案为120
为您推荐：
其他类似问题
由a1+a2+…+a100=200可知，s100=1+a100)&2=1+198)2=200，求出a1得到an的通项，a5+a10+a15+…+a100为等差数列，求出首项和公差即可求出之和．
本题考点：
等差数列的前n项和；等差数列的性质．
考点点评：
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．
a=-97a5=a+4*2=-89 a5,a10,……以-89为首项，2*5=10为公差的等差数列 20项 ...a5+a10+a15+……+a100 =-89*20+(19*20/2)*10 =120
扫描下载二维码>> MRPA151131FDT2A100 pdf Tama Electric
MRPA151131FDT2A100 价格
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型号：MRPA151131FDT2A100
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制造商：Tama Electric
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MRPA151131FDT2A100 PDF下载
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