2019年高考数学总复习课件：第五章单元检测-(共21张PPT)

第五章单元检测【答案】B一、选择题(每小题5分，共75分)1.252,5,22,11,...A.6B.7C.8D.9是数列中的第项【答案】C2.a,b,c都是实数,那么“2b=a+c”是“a,b,c成等差数列”的()A.充分非必要条件B.必要非充分条件C.充要条件D.既非充分又非必要条件【答案】A512311113.{},24,27,3A.,2B.2,2233C.,2D.,222naaaaaaqaqaqaq在等比数列中且则有【答案】C4.在等差数列{an}中,a2+a5=19,S5=40,则a10为()A.27B.28C.29D.30【答案】D5.若三个连续正整数的和是27,则在它前面的三个连续正整数的积是()A.120B.720C.72D.210【答案】C6.在首项为31,公差为-4的等差数列中,与0最靠近的项是()A.第7项B.第8项C.第9项D.第10项【答案】A7.在等比数列{an}中,a1+a2=10,a3+a4=40,则a5+a6等于()A.160B.±160C.70D.±70【答案】A6938.{},6,9,316A.4B.C.D.229naaaa在等比数列中如果那么为【答案】A9.在等差数列{an}中,若a3+a9+a15+a17=4,则a11等于()A.1B.-1C.0D.不确定【答案】B10.在等比数列{an}中,若a3a5=5,则a1a3a5a7=()A.10B.25C.50D.75【答案】A11.,,2,44A.1B.2C.0D.3aaaa已知为实数且成等比数列,则【答案】C37534612.1{},,A.B.C.D.nqaapapqpqpqpq公比的等比数列中若则【答案】B13.三个数成等差数列,其和为9,它们依次加上1,3,13就成等比数列,则这三个数为()A.0,3,6B.1,3,5C.5,3,1D.6,3,0【答案】B14.等比数列的前5项之和是10,前10项之和是30,则前15项之和是()A.60B.70C.0D.126【答案】B215.{}=32,A.65B.61C.32D.23nnnanSnnannnn已知数列的前项和为则二、填空题(每小题5分,共25分)16.已知数列通项为an=(-1)n+2n,则a25+a30=.17.已知(x+1)+(x+3)+(x+5)+…+(x+15)=96,则x=.18.已知c≠0,且a,b,c,2b成等差数列,则=.19.已知等差数列{an}中,a7+a9=16,a4=1,则a12=.20.在等比数列{an}中,an0,若S3=6,a7+a8+a9=24,则S30=.ac110415613813三、解答题(共50分)3214121111(1)188429322,2727()3322-,27(27)()33nnnnaaqaaqqqqaaqaa【解】①②由①②得当时当时243452621.{},(1)18,8,;(5)1(2),.(5)8nnaaaaaaaaa在等比数列中若已知求分若已知求的值分33454422264111(2)88211()24aaaaaaaa123222112115,315,5,-2,(1)21,(1)(1)322.22nnaaaaadaaaandnnnnnSnadnnn【解】在等差数列中所以即从而所以112322.{},3,15,.(12)nnnaaaaaanS已知数列是等差数列且求数列的通项和前项和分123111123.{},2,4,6,...,2,39273.(14)nnnnaaaaanS已知数列求分1211323132...111(2)(4)...(2)393111(24...2)(...)393(1)(22)112122311.223nnnnnnnSaaannnnnnnn【解】1111112(1)(),211112(-)-,1,211{}1,;21112(2)(1)1(1),.221nnnnnnnnnnnnnnaafaaaaaaaaaaannaan【解】由得即有且故是首项为公差为的等差数列由得*111224.()(-2),1,(),,N.21(1){};(2){}.(14)nnnnnnnxfxxaafabaanxaa设令又证明是等差数列求数列的通项公式分