1、一、选择题1.在等差数列an中,a13a3a1510,则a5的值为()A.2 B.3 C.4 D.5解析设数列an的公差为d,a1a152a8,2a83a310,2(a53d)3(a52d)10,5a510,a52.答案A2.设等比数列an的前n项和为Sn,若Sm15,Sm11,Sm121,则m等于()A.3 B.4 C.5 D.6解析由已知得SmSm1am16,Sm1Smam132,故公比q2,又Sm11,故a11,又ama1qm116,代入可求得m5.答案C3.等差数列an的公差为2,若a2,a4,a8成等比数列,则an的前n项和Sn等于()A.n(n1) B.n(n1)C. D.解析由a
2、2,a4,a8成等比数列,得aa2a8,即(a16)2(a12)(a114),a12.Sn2n22nn2nn(n1).答案A4.设各项都是正数的等比数列an,Sn为前n项和,且S1010,S3070,那么S40等于()A.150 B.200C.150或200 D.400或50解析依题意,数列S10,S20S10,S30S20,S40S30成等比数列,因此有(S20S10)2S10(S30S20),即(S2010)210(70S20),故S2020或S2030.又S200,因此S2030,S20S1020,S30S2040,则S40S3070150.答案A5.(2015浙江卷)已知an是等差数列
3、,公差d不为零,前n项和是Sn,若a3,a4,a8成等比数列,则()A.a1d0,dS40 B.a1d0,dS40C.a1d0,dS40 D.a1d0,dS40解析a3,a4,a8成等比数列,(a13d)2(a12d)(a17d),整理得a1d,a1dd20,又S44a1d,dS40,故选B.答案B二、填空题6.(2016江苏卷)已知an是等差数列,Sn是其前n项和.若a1a3,S510,则a9的值是_.解析设等差数列an公差为d,由题意可得:解得则a9a18d48320.答案207.若数列an的前n项和Snan,则an的通项公式是_.解析当n2时,Sn1an1,anSnSn1anan1ana
4、n1.an2an1,又n1时,a11,数列an是以1为首项,2为公比的等比数列,所以an(2)n1.答案an(2)n18.(2015全国卷)设Sn是数列an的前n项和,且a11,an1SnSn1,则Sn_.解析由题意,得S1a11,又由an1SnSn1,得Sn1SnSnSn1,所以Sn0,所以1,即1,故数列是以1为首项,1为公差的等差数列,得1(n1)n,所以Sn.答案三、解答题9.(2016全国卷)已知an是公差为3的等差数列,数列bn满足b11,b2,anbn1bn1nbn.(1)求an的通项公式;(2)求bn的前n项和.解(1)由已知,a1b2b2b1,b11,b2,得a12.所以数列
5、an是首项为2,公差为3的等差数列,通项公式为an3n1.(2)由(1)和anbn1bn1nbn得bn1,因此bn是首项为1,公比为的等比数列.记bn的前n项和为Sn,则Sn.10.(2016洛阳模拟)已知数列an满足a11,an13an1,(1)证明是等比数列,并求an的通项公式;(2)证明.证明(1)由an13an1,得an13.又a1,所以是首项为,公比为3的等比数列.an,因此an的通项公式为an.(2)由(1)知.因为当n1时,3n123n1,所以.于是1.所以.11.(2016泉州二模)已知等比数列an满足:|a2a3|10,a1a2a3125.(1)求数列an的通项公式;(2)是否存在正整数m,使得1?若存在,求m的最小值;若不存在,说明理由.解(1)设等比数列an的公比为q,则由已知可得解得或故an3n1或an5(1)n1.(2)若an3n1,则,则是首项为,公比为的等比数列.从而1.若an5(1)n1,则(1)n1,故是首项为,公比为1的等比数列,从而故1.综上,对任何正整数m,总有1.故不存在正整数m,使得1成立.