2010-2019十年高考數(shù)學(xué)真題分類匯編專題08 數(shù)列學(xué)生版+解析版

1、十年高考真題分類匯編（20102019）數(shù)學(xué)專題08數(shù)列一、選擇題1.(2019全國(guó)1理t9)記sn為等差數(shù)列an的前n項(xiàng)和.已知s4=0,a5=5,則()a.an=2n-5c.sn=2n2-8nb.an=3n-102d.sn=1n2-2n22.(2019浙江t10)設(shè)a,br,數(shù)列an滿足a1=a,an+1=+b,nn*,則()2a.當(dāng)b=1時(shí),a1010c.當(dāng)b=-2時(shí),a10104b.當(dāng)b=1時(shí),a1010d.當(dāng)b=-4時(shí),a10103.(2018全國(guó)1理t4)記sn為等差數(shù)列an的前n項(xiàng)和,若3s3=s2+s4,a1=2,則a5=()a.-12b.-10c.10d.124.(2018浙

2、江t10)已知a1,a2,a3,a4成等比數(shù)列,且a1+a2+a3+a4=ln(a1+a2+a3).若a11,則()a.a1a3,a2a3,a2a4c.a1a4d.a1a3,a2a45.(2018北京理t4文t5)“十二平均律”是通用的音律體系,明代朱載堉最早用數(shù)學(xué)方法計(jì)算出半音比例,為這個(gè)理論的發(fā)展做出了重要貢獻(xiàn).十二平均律將一個(gè)純八度音程分成十二份,依次得到十三個(gè)單音,從第二個(gè)單音起,每一個(gè)單音的頻率與它的前一個(gè)單音的頻率的比都等于122.若第一個(gè)單音的頻率為f,則第八個(gè)單音的頻率為()3a.2f3b.22fc.1225fd.1227f6.(2017全國(guó)1理t12)幾位大學(xué)生響應(yīng)國(guó)家的創(chuàng)業(yè)

3、號(hào)召,開發(fā)了一款應(yīng)用軟件.為激發(fā)大家學(xué)習(xí)數(shù)學(xué)的興趣,他們推出了“解數(shù)學(xué)題獲取軟件激活碼”的活動(dòng).這款軟件的激活碼為下面數(shù)學(xué)問(wèn)題的答案:已知數(shù)列1,1,2,1,2,4,1,2,4,8,1,2,4,8,16,其中第一項(xiàng)是20,接下來(lái)的兩項(xiàng)是20,21,再接下來(lái)的三項(xiàng)是20,21,22,依此類推.求滿足如下條件的最小整數(shù)n:n100且該數(shù)列的前n項(xiàng)和為2的整數(shù)冪.那么該款軟件的激活碼是()a.440b.330c.220d.1107.(2017全國(guó)3理t9)等差數(shù)列an的首項(xiàng)為1,公差不為0.若a2,a3,a6成等比數(shù)列,則an前6項(xiàng)的和為()a.-24b.-3c.3d.88.(2016全國(guó)1理t3)

4、已知等差數(shù)列an前9項(xiàng)的和為27,a10=8,則a100=()1a.100b.99c.98d.979.(2015浙江理t13)已知an是等差數(shù)列,公差d不為零,前n項(xiàng)和是sn,若a3,a4,a8成等比數(shù)列,則()a.a1d0,ds40b.a1d0,ds40,ds40d.a1d010.(2015全國(guó)2文t5)設(shè)sn是等差數(shù)列an的前n項(xiàng)和,若a1+a3+a5=3,則s5=()a.5b.7c.9d.112b.192c.10d.122d.12d.(-1)3b.-13c.19d.-111.(2015全國(guó)1文t7)已知an是公差為1的等差數(shù)列,sn為an的前n項(xiàng)和.若s8=4s4,則a10=(a.171

5、2.(2015全國(guó)2理t4)已知等比數(shù)列an滿足a1=3,a1+a3+a5=21,則a3+a5+a7=()a.21b.42c.63d.84413.(2015全國(guó)2文t9)已知等比數(shù)列an滿足a1=1,a3a5=4(a4-1),則a2=()a.2b.1c.1814.(2014大綱全國(guó)文t8)設(shè)等比數(shù)列an的前n項(xiàng)和為sn.若s2=3,s4=15,則s6=()a.31b.32c.63d.6415.(2014全國(guó)2文t5)等差數(shù)列an的公差為2,若a2,a4,a8成等比數(shù)列,則an的前n項(xiàng)和sn=(a.n(n+1)b.n(n-1)c.(+1)216.(2013全國(guó)2理t3)等比數(shù)列an的前n項(xiàng)和為s

6、n.已知s3=a2+10a1,a5=9,則a1=()a.19317.(2013全國(guó)1文t6)設(shè)首項(xiàng)為1,公比為2的等比數(shù)列an的前n項(xiàng)和為sn,則()a.sn=2an-1b.sn=3an-2c.sn=4-3and.sn=3-2an18.(2013全國(guó)1理t12)設(shè)abncn的三邊長(zhǎng)分別為an,bn,c,anbncn的面積為sn,n=1,2,3,.若22b1c1,b1+c1=2a1,an+1=an,bn+1=cn+an,cn+1=bn+an,則()a.sn為遞減數(shù)列b.sn為遞增數(shù)列c.s2n-1為遞增數(shù)列,s2n為遞減數(shù)列)2d.s2n-1為遞減數(shù)列,s2n為遞增數(shù)列19.(2013全國(guó)1理t

7、7)設(shè)等差數(shù)列an的前n項(xiàng)和為sn,若sm-1=-2,sm=0,sm+1=3,則m=()a.3b.4c.5d.620.(2012全國(guó)理t5)已知an為等比數(shù)列,a4+a7=2,a5a6=-8,則a1+a10=()a.7b.5c.-5d.-721.(2012全國(guó)文t12)數(shù)列an滿足an+1+(-1)nan=2n-1,則an的前60項(xiàng)和為()a.3690b.3660c.1845d.18302.(2019全國(guó)3理t14)記sn為等差數(shù)列an的前n項(xiàng)和.若a10,a2=3a1,則s10=.二、填空題1.(2019全國(guó)3文t14)記sn為等差數(shù)列an的前n項(xiàng)和.若a3=5,a7=13,則s10=.s5

8、3.(2019江蘇t8)已知數(shù)列an(nn*)是等差數(shù)列,sn是其前n項(xiàng)和.若a2a5+a8=0,s9=27,則s8的值是.4.(2019北京理t10)設(shè)等差數(shù)列an的前n項(xiàng)和為sn.若a2=-3,s5=-10,則a5=,sn的最小值為.4345.(2019全國(guó)1文t14)記sn為等比數(shù)列an的前n項(xiàng)和.若a1=1,s3=3,則s4=.6.(2019全國(guó)1理t14)記sn為等比數(shù)列an的前n項(xiàng)和.若a1=1,2=a6,則s5=_.7.(2018全國(guó)1理t14)記sn為數(shù)列an的前n項(xiàng)和.若sn=2an+1,則s6=.8.(2018北京理t9)設(shè)an是等差數(shù)列,且a1=3,a2+a5=36,則a

9、n的通項(xiàng)公式為.19.(2018上海t10)設(shè)等比數(shù)列an的通項(xiàng)公式為an=qn-1(nn*),前n項(xiàng)和為sn,若limasn=2,則q=.nn+110.(2018江蘇t14)已知集合a=x|x=2n-1,nn*,b=x|x=2n,nn*.將ab的所有元素從小到大依次排列構(gòu)成一個(gè)數(shù)列an.記sn為數(shù)列an的前n項(xiàng)和,則使得sn12an+1成立的n的最小值為.11.(2017全國(guó)2理t15)等差數(shù)列an的前n項(xiàng)和為sn,a3=3,s4=10,則1=_.=112.(2017全國(guó)3理t14)設(shè)等比數(shù)列an滿足a1+a2=-1,a1-a3=-3,則a4=.4413.(2017江蘇理t9文t9)等比數(shù)列

10、an的各項(xiàng)均為實(shí)數(shù),其前n項(xiàng)和為sn.已知s3=7,s6=63,則a8=.14.(2016浙江理t13文t13)設(shè)數(shù)列an的前n項(xiàng)和為sn,若s2=4,an+1=2sn+1,nn*,則a1=,s5=.15.(2016北京理t12)已知an為等差數(shù)列,sn為其前n項(xiàng)和.若a1=6,a3+a5=0,則s6=.16.(2016全國(guó)1理t15)設(shè)等比數(shù)列an滿足a1+a3=10,a2+a4=5,則a1a2an的最大值為.17.(2015全國(guó)1文t13)在數(shù)列an中,a1=2,an+1=2an,sn為an的前n項(xiàng)和.若sn=126,則n=.18.(2015湖南理t14)設(shè)sn為等比數(shù)列an的前n項(xiàng)和,若

11、a1=1,且3s1,2s2,s3成等差數(shù)列,則an=.19.(2015福建文t16)若a,b是函數(shù)f(x)=x2-px+q(p0,q0)的兩個(gè)不同的零點(diǎn),且a,b,-2這三個(gè)數(shù)可適3an=n+1(nn).則數(shù)列前10項(xiàng)的和為_.*127.(2014全國(guó)2文t16)數(shù)列an滿足an+1=,a8=2,則a1=_.1當(dāng)排序后成等差數(shù)列,也可適當(dāng)排序后成等比數(shù)列,則p+q的值等于.20.(2015江蘇理t11)設(shè)數(shù)列an滿足a1=1,且an+1-21.(2015全國(guó)2理t16)設(shè)sn是數(shù)列an的前n項(xiàng)和,且a1=-1,an+1=snsn+1,則sn=.22.(2015廣東理t10)在等差數(shù)列an中,若

12、a3+a4+a5+a6+a7=25,則a2+a8=.23.(2015陜西文t13)中位數(shù)為1010的一組數(shù)構(gòu)成等差數(shù)列,其末項(xiàng)為2015,則該數(shù)列的首項(xiàng)為.24.(2014江蘇理t7)在各項(xiàng)均為正數(shù)的等比數(shù)列an中,若a2=1,a8=a6+2a4,則a6的值是.25.(2014廣東文t13)等比數(shù)列an的各項(xiàng)均為正數(shù),且a1a5=4,則log2a1+log2a2+log2a3+log2a4+log2a5=.26.(2014安徽理t12)數(shù)列an是等差數(shù)列,若a1+1,a3+3,a5+5構(gòu)成公比為q的等比數(shù)列,則q=.1-an28.(2014北京理t12)若等差數(shù)列an滿足a7+a8+a90,a

13、7+a100,則當(dāng)n=時(shí),an的前n項(xiàng)和最大.29.(2014天津理t11)設(shè)an是首項(xiàng)為a1,公差為-1的等差數(shù)列,sn為其前n項(xiàng)和.若s1,s2,s4成等比數(shù)列,則a1的值為.30.(2013全國(guó)2理t16)等差數(shù)列an的前n項(xiàng)和為sn,已知s10=0,s15=25,則nsn的最小值為.31.(2013遼寧理t14)已知等比數(shù)列an是遞增數(shù)列,sn是an的前n項(xiàng)和.若a1,a3是方程x2-5x+4=0的兩個(gè)根,則s6=.3332.(2013全國(guó)1理t14)若數(shù)列an的前n項(xiàng)和sn=2an+1,則an的通項(xiàng)公式是an=.33.(2012全國(guó)文t14)等比數(shù)列an的前n項(xiàng)和為sn,若s3+3s

14、2=0,則公比q=.三、計(jì)算題1.(2019全國(guó)2文t18)已知an是各項(xiàng)均為正數(shù)的等比數(shù)列,a1=2,a3=2a2+16.(1)求an的通項(xiàng)公式;(2)設(shè)bn=log2an.求數(shù)列bn的前n項(xiàng)和.2.(2019全國(guó)2理t19)已知數(shù)列an和bn滿足a1=1,b1=0,4an+1=3an-bn+4,4bn+1=3bn-an-4.(1)證明:an+bn是等比數(shù)列,an-bn是等差數(shù)列;(2)求an和bn的通項(xiàng)公式.3.(2019天津文t18)設(shè)an是等差數(shù)列,bn是等比數(shù)列,公比大于0.已知a1=b1=3,b2=a3,b3=4a2+3.(1)求an和bn的通項(xiàng)公式;4,=2,(2)記cn=*,n

15、n,證明:c1+c2+cn2n,nn.(2)已知數(shù)列bn(nn)滿足:b1=1,*1=22,其中sn為數(shù)列bn的前n項(xiàng)和.,為偶數(shù),(2)設(shè)數(shù)列cn滿足cn=1,為奇數(shù),求a1c1+a2c2+a2nc2n(nn*).24.(2019天津理t19)設(shè)an是等差數(shù)列,bn是等比數(shù)列.已知a1=4,b1=6,b2=2a2-2,b3=2a3+4.(1)求an和bn的通項(xiàng)公式;1,20,mn*,q(1,2,證明:存在dr,使得|an-bn|b1對(duì)n=2,3,m+1均成立,并求d的取值范圍(用b1,m,q表示).10.(2018天津文t18)設(shè)an是等差數(shù)列,其前n項(xiàng)和為sn(nn*);bn是等比數(shù)列,公

16、比大于0,其前n項(xiàng)和為tn(nn*).已知b1=1,b3=b2+2,b4=a3+a5,b5=a4+2a6.(1)求sn和tn;(2)若sn+(t1+t2+tn)=an+4bn,求正整數(shù)n的值.11.(2018天津理t18)設(shè)an是等比數(shù)列,公比大于0,其前n項(xiàng)和為sn(nn*),bn是等差數(shù)列.已知a1=1,a3=a2+2,a4=b3+b5,a5=b4+2b6.(1)求an和bn的通項(xiàng)公式;(2)設(shè)數(shù)列sn的前n項(xiàng)和為tn(nn*),求tn;(+1)(+2)-2(nn*).1證明(+2)2+2+212.(2018全國(guó)2理t17文t17)記sn為等差數(shù)列an的前n項(xiàng)和,已知a1=-7,s3=-1

17、5.(1)求an的通項(xiàng)公式;(2)求sn,并求sn的最小值.13.(2018全國(guó)1文t17)已知數(shù)列an滿足a1=1,nan+1=2(n+1)an.設(shè)bn=.(1)求b1,b2,b3;(2)判斷數(shù)列bn是否為等比數(shù)列,并說(shuō)明理由;(3)求an的通項(xiàng)公式.14.(2018全國(guó)3理t17文t17)等比數(shù)列an中,a1=1,a5=4a3.(1)求an的通項(xiàng)公式;(2)記sn為an的前n項(xiàng)和,若sm=63,求m.15.(2017全國(guó)1文t17)設(shè)sn為等比數(shù)列an的前n項(xiàng)和,已知s2=2,s3=-6.(1)求an的通項(xiàng)公式;(2)求sn,并判斷sn+1,sn,sn+2是否成等差數(shù)列.16.(2017全

18、國(guó)2文t17)已知等差數(shù)列an的前n項(xiàng)和為sn,等比數(shù)列bn的前n項(xiàng)和為tn,a1=-1,b1=1,a2+b2=2.(1)若a3+b3=5,求bn的通項(xiàng)公式;6(2)若t3=21,求s3.17.(2017全國(guó)3文t17)設(shè)數(shù)列an滿足a1+3a2+(2n-1)an=2n.(1)求an的通項(xiàng)公式;(2)求數(shù)列的前n項(xiàng)和.2+118.(2017天津理t18)已知an為等差數(shù)列,前n項(xiàng)和為sn(nn*),bn是首項(xiàng)為2的等比數(shù)列,且公比大于0,b2+b3=12,b3=a4-2a1,s11=11b4.(1)求an和bn的通項(xiàng)公式;(2)求數(shù)列a2nb2n-1的前n項(xiàng)和(nn*).19.(2017山東理

19、t19)已知xn是各項(xiàng)均為正數(shù)的等比數(shù)列,且x1+x2=3,x3-x2=2.(1)求數(shù)列xn的通項(xiàng)公式;(2)如圖,在平面直角坐標(biāo)系xoy中,依次連接點(diǎn)p1(x1,1),p2(x2,2)pn+1(xn+1,n+1)得到折線p1p2pn+1,求由該折線與直線y=0,x=x1,x=xn+1所圍成的區(qū)域的面積tn.(2)bn為各項(xiàng)非零的等差數(shù)列,其前n項(xiàng)和為sn.已知s2n+1=bnbn+1,求數(shù)列的前n項(xiàng)和tn.20.(2017山東文t19)已知an是各項(xiàng)均為正數(shù)的等比數(shù)列,且a1+a2=6,a1a2=a3.1)求數(shù)列an的通項(xiàng)公式;21.(2017天津文t18)已知an為等差數(shù)列,前n項(xiàng)和為sn

20、(nn*),bn是首項(xiàng)為2的等比數(shù)列,且公比大于0,b2+b3=12,b3=a4-2a1,s11=11b4.(1)求an和bn的通項(xiàng)公式;(2)求數(shù)列a2nbn的前n項(xiàng)和(nn*).22.(2016全國(guó)2理t17)sn為等差數(shù)列an的前n項(xiàng)和,且a1=1,s7=28.記bn=lgan,其中x表示不超過(guò)x的最大整數(shù),如0.9=0,lg99=1.(1)求b1,b11,b101;(2)求數(shù)列bn的前1000項(xiàng)和.7(2)令cn=(+1)(+2),求數(shù)列cn的前n項(xiàng)和tn.(1)設(shè)cn=+12,nn*,求證:數(shù)列cn是等差數(shù)列;23.(2016全國(guó)2文t17)等差數(shù)列an中,a3+a4=4,a5+a7

21、=6.(1)求an的通項(xiàng)公式;(2)設(shè)bn=an,求數(shù)列bn的前10項(xiàng)和,其中x表示不超過(guò)x的最大整數(shù),如0.9=0,2.6=2.24.(2016浙江文t17)設(shè)數(shù)列an的前n項(xiàng)和為sn.已知s2=4,an+1=2sn+1,nn*.(1)求通項(xiàng)公式an;(2)求數(shù)列|an-n-2|的前n項(xiàng)和.25.(2016北京文t15)已知an是等差數(shù)列,bn是等比數(shù)列,且b2=3,b3=9,a1=b1,a14=b4.(1)求an的通項(xiàng)公式;(2)設(shè)cn=an+bn,求數(shù)列cn的前n項(xiàng)和.26.(2016山東理t18文t19)已知數(shù)列an的前n項(xiàng)和sn=3n2+8n,bn是等差數(shù)列,且an=bn+bn+1.

22、(1)求數(shù)列bn的通項(xiàng)公式;+127.(2016天津理t18)已知an是各項(xiàng)均為正數(shù)的等差數(shù)列,公差為d.對(duì)任意的nn*,bn是an和an+1的等比中項(xiàng).2(2)設(shè)a1=d,tn=(-1)k228.(2016天津文t18)已知an是等比數(shù)列,前n項(xiàng)和為sn(nn*),且1122,nn*,求證:1110,+2an=4sn+3.(1)求an的通項(xiàng)公式;1+136.(2015安徽文t18)已知數(shù)列an是遞增的等比數(shù)列,且a1+a4=9,a2a3=8.(1)求數(shù)列an的通項(xiàng)公式;+137.(2015天津理t18)已知數(shù)列an滿足an+2=qan(q為實(shí)數(shù),且q1),nn*,a1=1,a2=2,且a2+

23、a3,a3+a4,a4+a5成等差數(shù)列.(1)求q的值和an的通項(xiàng)公式;222-112+1(1)求數(shù)列an的通項(xiàng)公式;(2)設(shè)bn=(an+1)2,求數(shù)列bn的前n項(xiàng)和tn.23139.(2015浙江文t17)已知數(shù)列an和bn滿足a1=2,b1=1,an+1=2an(nn*),b1+1b2+1b3+bn=bn+1-1(nn*).(1)求an與bn;(2)記數(shù)列anbn的前n項(xiàng)和為tn,求tn.40.(2015天津文t18)已知an是各項(xiàng)均為正數(shù)的等比數(shù)列,bn是等差數(shù)列,且a1=b1=1,b2+b3=2a3,a5-3b2=7.(1)求an和bn的通項(xiàng)公式;9(2)當(dāng)d1時(shí),記cn=,求數(shù)列c

24、n的前n項(xiàng)和tn.(2)設(shè)cn=anbn,nn*,求數(shù)列cn的前n項(xiàng)和.41.(2015湖北文t19)設(shè)等差數(shù)列an的公差為d,前n項(xiàng)和為sn,等比數(shù)列bn的公比為q,已知b1=a1,b2=2,q=d,s10=100.(1)求數(shù)列an,bn的通項(xiàng)公式;42.(2014全國(guó)2理t17)已知數(shù)列an滿足a1=1,an+1=3an+1.2(1)證明:+1是等比數(shù)列,并求an的通項(xiàng)公式;(2)證明:211+1+110c.當(dāng)b=-2時(shí),a10101b.當(dāng)b=4時(shí),a1010d.當(dāng)b=-4時(shí),a1010【答案】a2222222232162【解析】當(dāng)b=1時(shí),a2=1+11,a3=2+14,a4=3+117

25、1,當(dāng)n4時(shí),an+1=+11,則16lo17an+12lo17anlo17an+12n-1,則an+1（17）1616162-11616(n4),則a10（17）26=（1+16）64=1+64+1+1+4+710,故選a.646321623.(2018全國(guó)1理t4)記sn為等差數(shù)列an的前n項(xiàng)和,若3s3=s2+s4,a1=2,則a5=()a.-12b.-10c.10d.12a1+a2+a3+a4=1(1-),a1+a2+a3=1(1-).【答案】b【解析】因?yàn)?s3=s2+s4,所以3s3=(s3-a3)+(s3+a4),即s3=a4-a3.設(shè)公差為d,則3a1+3d=d,又由a1=2,

26、得d=-3,所以a5=a1+4d=-10.4.(2018浙江t10)已知a1,a2,a3,a4成等比數(shù)列,且a1+a2+a3+a4=ln(a1+a2+a3).若a11,則()a.a1a3,a2a3,a2a4c.a1a4d.a1a3,a2a4【答案】b【解析】設(shè)等比數(shù)列的公比為q,則431-1-13a1+a2+a3+a4=ln(a1+a2+a3),a1+a2+a3=1234,即a1(1+q+q2)=1(123).又a11,q1,即q+q20,解得q0舍去).由a11,可知a1(1+q+q2)1,a1(1+q+q2+q3)0,即1+q+q2+q30,即(1+q)+q2(1+q)0,即(1+q)(1

27、+q2)0,這與q-1相矛盾.1+q+q21,即-1qa3,a2100且該數(shù)列的前n項(xiàng)和為2的整數(shù)冪.那么該款軟件的激活碼是()a.440b.330c.220d.110【答案】a【解析】設(shè)數(shù)列的首項(xiàng)為第1組,接下來(lái)兩項(xiàng)為第2組,再接下來(lái)三項(xiàng)為第3組,以此類推,設(shè)第n組的項(xiàng)數(shù)21-21-2由題意,n100,令(1)100,得n14且nn*,即n出現(xiàn)在第13組之后.若要使最小整數(shù)n滿足:n100且2前n項(xiàng)和為2的整數(shù)冪,則sn-(1)應(yīng)與-2-n互為相反數(shù),即2k-1=2+n(kn*,n14),所以k=log2(n+3),解2得n=29,k=5.14所以n=29(1+29)+5=440,故選a.2

28、7.(2017全國(guó)3理t9)等差數(shù)列an的首項(xiàng)為1,公差不為0.若a2,a3,a6成等比數(shù)列,則an前6項(xiàng)的和為()a.-24b.-3c.3d.8【答案】a【解析】設(shè)等差數(shù)列的公差為d,則d0,a23=a2a6,即(1+2d)2=(1+d)(1+5d),解得d=-2,2所以s6=61+65(-2)=-24,故選a.8.(2016全國(guó)1理t3)已知等差數(shù)列an前9項(xiàng)的和為27,a10=8,則a100=()a.100b.99c.98d.97【答案】c2-a【解析】因?yàn)閟9=(a1+a9)9=27,a1+a9=2a5,所以a5=3.又因?yàn)閍10=8,所以d=a10-55=1.故a100=a10+(1

29、00-10)1=98.9.(2015浙江理t13)已知an是等差數(shù)列,公差d不為零,前n項(xiàng)和是sn,若a3,a4,a8成等比數(shù)列,則(a.a1d0,ds40b.a1d0,ds40,ds40d.a1d0【答案】b【解析】設(shè)an的首項(xiàng)為a1,公差為d,則a3=a1+2d,a4=a1+3d,a8=a1+7d.a3,a4,a8成等比數(shù)列,(a1+3d)2=(a1+2d)(a1+7d),即3a1d+5d2=0.33d0,a1d=-5d20,且a1=-5d.13ds4=4d(a2+a4)=2d(2a1+3d)=-2d20.10.(2015全國(guó)2文t5)設(shè)sn是等差數(shù)列an的前n項(xiàng)和,若a1+a3+a5=3,則s5=()a.5b.7c.9d.11【答案】a【解析】由a1+a3+a5=3及等差中項(xiàng),得3a3=3,解得a3=1.故)152b.192c.10d.12【解析】由題意知1+3+5=1+q2+q4=21=7,解得q2=2(負(fù)值舍去).a3+a5+a7=(a1+a