全国通用2016高考数学二轮复习专题三第1讲等差数列等比数列训练文.doc

15第1讲等差数列、等比数列一、选择题1.(2015·广州模拟)在等差数列{an}中a1＋3a3＋a15＝10则a5的值为()A.2B.3C.4D.5解析设数列{an}的公差为d∵a1＋a15＝2a8∴2a8＋3a3＝10∴2(a5＋3d)＋3(a5－2d)＝10∴5a5＝10∴a5＝2.答案A2.(2015·济南模拟)等比数列{an}的前n项和为Sn若2S4＝S5＋S6则数列{an}的公比q的值为()A.－2或1B.－1或2C.－2D.1解析法一若q＝1则S4＝4a1S5＝5a1S6＝6a1显然不满足2S4＝S5＋S6故A、D错.若q＝－1则S4＝S6＝0S5＝a5≠0不满足条件故B错因此选C.法二经检验q＝1不适合则由2S4＝S5＋S6得2(1－q4)＝1－q5＋1－q6化简得q2＋q－2＝0解得q＝1(舍去)q＝－2.答案C3.已知{an}为等差数列其公差为－2且a7是a3与a9的等比中项Sn为{an}的前n项和n∈N*则S10的值为()A.－110B.－90C.90D.110解析∵a3＝a1＋2d＝a1－4a7＝a1＋6d＝a1－12a9＝a1＋8d＝a1－16又∵a7是a3与a9的等比中项∴(a1－12)2＝(a1－4)·(a1－16)解得a1＝20.∴S10＝10×20＋eq\f(12)×10×9×(－2)＝110.答案D4.(2014·新课标全国Ⅱ卷)等差数列{an}的公差为2若a2a4a8成等比数列则{an}的前n项和Sn等于()A.n(n＋1)B.n(n－1)C.eq\f(n（n＋1）2)D.eq\f(n（n－1）2)解析由a2a4a8成等比数列得aeq\o\al(24)＝a2a8即(a1＋6)2＝(a1＋2)(a1＋14)∴a1＝2.∴Sn＝2n＋eq\f(n（n－1）2)×2＝2n＋n2－n＝n(n＋1).答案A5.(2015·南昌模拟)在正项等比数列{an}中3a1eq\f(12)a32a2成等差数列则eq\f(a2014＋a2015a2012＋a2013)等于()A.3或－1B.9或1C.1D.9解析依题意有3a1＋2a2＝a3即3a1＋2a1q＝a1q2解得q＝3eq\f(a2014＋a2015a2012＋a2013)＝eq\f(a1q2013＋a1q2014a1q2011＋a1q2012)＝eq\f(q2＋q31＋q)＝9.答案D二、填空题6.在等比数列{an}中a1＝3a4＝24则a3＋a4＋a5等于________.解析∵a4＝a1q3∴q3＝8∴q＝2∴a3＋a4＋a5＝a1q2(1＋q＋q2)＝84.答案847.(2015·阳泉模拟)若等差数列{an}满足a7＋a8＋a9＞0a7＋a10＜0则当n＝________时{an}的前n项和最大.解析根据题意知a7＋a8＋a9＝3a8＞0即a8＞0.又a8＋a9＝a7＋a10＜0∴a9＜0∴当n＝8时{an}的前n项和最大.答案88.等比数列{an}的各项均为正数且a1a5＝4则log2a1＋log2a2＋log2a3＋log2a4＋log2a5＝________.解析由等比数列的性质可知a1a5＝a2a4＝aeq\o\al(23)于是由a1a5＝4得a3＝2故a1a2a3a4a5＝32则log2a1＋log2a2＋log2a3＋log2a4＋log2a5＝log2(a1a2a3a4a5)＝log232＝5.答案5三、解答题9.(2015·重庆卷)已知等差数列{an}满足a3＝2前3项和S3＝eq\f(92).(1)求{an}的通项公式；(2)设等比数列{bn}满足b1＝a1b4＝a15求{bn}的前n项和Tn.解(1)设{an}的公差为d则由已知条件得a1＋2d＝23a1＋eq\f(3×22)d＝eq\f(92)化简得a1＋2d＝2a1＋d＝eq\f(32)解得a1＝1d＝eq\f(12)故通项公式an＝1＋eq\f(n－12)即an＝eq\f(n＋12).(2)由(1)得b1＝1b4＝a15＝eq\f(15＋12)＝8.设{bn}的公比为q则q3＝eq\f(b4b1)＝8从而q＝2故{