（適用于新高考新教材）高考數(shù)學(xué)一輪總復(fù)習(xí)第六章數(shù)列課時(shí)規(guī)范練30等差數(shù)列

課時(shí)規(guī)范練30等差數(shù)列基礎(chǔ)鞏固組1.(2022廣東廣州二模)已知數(shù)列{an}是等差數(shù)列,且a2+a5+a8=π,則tan(a1+a9)=()A.3 B.3C.33 D.2.在數(shù)列{an}中,a1=3,am+n=am+an(m,n∈N*),若a1+a2+a3+…+ak=135,則k=()A.10 B.9 C.8 D.73.在等差數(shù)列{an}中,a1與a4041是f(x)=x4lnxmx的兩個(gè)極值點(diǎn),則log2a2021=(A.1 B.2 C.0 D.14.已知數(shù)列{bn}是公差不為0的等差數(shù)列,且b1224b12=b201024b2010,則數(shù)列{bA.20214 B.2021C.2021 D.40425.已知兩個(gè)等差數(shù)列{an}和{bn}的前n項(xiàng)和分別為Sn和Tn,且SnTn=2n+70A.4 B.5 C.6 D.76.(多選)已知等差數(shù)列{an}是遞增數(shù)列,a7=3a5,前n項(xiàng)和為Sn,下列選項(xiàng)正確的是()A.d>0B.a1<0C.當(dāng)n=5時(shí)Sn最小D.Sn>0時(shí)n的最小值為87.(多選)在數(shù)列{an}中,a1=1,an+an+1=3n,則下列說(shuō)法正確的是()A.a6=8 B.{a2n}是等差數(shù)列C.S20=300 D.a2na2n1=38.已知等差數(shù)列{an}的公差為2,且a1,a2,a3+1成等比數(shù)列,Sn是數(shù)列{an}的前n項(xiàng)和,則S9=.

9.(2022遼寧鞍山二模)設(shè)等差數(shù)列{an},{bn}的前n項(xiàng)和分別是Sn,Tn,若SnTn=2n綜合提升組10.若等差數(shù)列{an}的前n項(xiàng)和為Sn,則“S2020>0,S2021<0”是“a1010a1011<0”的()A.充要條件B.充分不必要條件C.必要不充分條件D.既不充分也不必要條件11.設(shè)Sn是等差數(shù)列{an}的前n項(xiàng)和,若S5S10=13A.19 B.7C.113 D.12.(多選)設(shè)等差數(shù)列{an}的前n項(xiàng)和為Sn,公差為d,a3=12,S12>0,a7<0,則下列說(shuō)法正確的是()A.d<0B.a6<0 C.247<d<D.Sn<0時(shí),n的最小值為1313.在等差數(shù)列{an}中,a1+a7=12,當(dāng)a32+a42+14.(2022山東濰坊一模)2022年北京冬奧會(huì)開(kāi)幕式始于二十四節(jié)氣倒計(jì)時(shí),它將中國(guó)人的物候文明、傳承久遠(yuǎn)的詩(shī)歌、現(xiàn)代生活的畫(huà)面和諧統(tǒng)一起來(lái).我國(guó)古人將一年分為二十四個(gè)節(jié)氣,如圖所示,相鄰兩個(gè)節(jié)氣的日晷長(zhǎng)變化量相同,冬至日晷長(zhǎng)最長(zhǎng),夏至日晷長(zhǎng)最短,周而復(fù)始.已知冬至日晷長(zhǎng)為13.5尺,芒種日晷長(zhǎng)為2.5尺,則一年中夏至到大雪的日晷長(zhǎng)的和為尺.

15.(2022山東聊城一模)設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,對(duì)于任意的n∈N*都有an+1=an+2,且S6=4a5.(1)求數(shù)列{an}的通項(xiàng)公式;(2)若數(shù)列{bn}滿(mǎn)足bn=Sncosnπ,求數(shù)列{bn}的前2n項(xiàng)和T2n.創(chuàng)新應(yīng)用組16.(多選)(2022江蘇蘇錫常鎮(zhèn)一模)記Sn為等差數(shù)列{an}的前n項(xiàng)和,則()A.S6=2S4S2B.S6=3(S4S2)C.S2n,S4nS2n,S6nS4n成等差數(shù)列D.S2

課時(shí)規(guī)范練30等差數(shù)列1.D解析:在等差數(shù)列{an}中,a2+a5+a8=π,則有3a5=π,即a5=π3,所以tan(a1+a9)=tan2a5=tan2π32.B解析:令m=1,由am+n=am+an可得an+1=a1+an,所以an+1an=3,所以數(shù)列{an}是首項(xiàng)為3,公差為3的等差數(shù)列,所以an=3+3(n1)=3n,所以a1+a2+a3+…+ak=k(a1+ak)2=k(3+3k)23.B解析:f'(x)=14x+mx2=x2-4x+mx2(x>0).因?yàn)閍1與a4041是f(x)=x4lnxmx的兩個(gè)極值點(diǎn),所以a1與a4041是方程x24x+m=0的兩個(gè)根,即a1+a4041=4,即2a2021=4,所以4.D解析:∵數(shù)列{bn}是公差不為0的等差數(shù)列,且b1224b12=b201024b2010,∴(b12b2010)(b12+b2010)=4(b12b2010),且b12≠b2010,∴b12∴數(shù)列{bn}的前2021項(xiàng)和S2021=20212(b1+b2021)=20212(b12+b2010)=20212×4=40425.B解析:依題意,anbn=S2n-1T2n-1=2(2n-1)+70(2n-16.ABD解析:由題意,設(shè)等差數(shù)列{an}的公差為d.因?yàn)閍7=3a5,所以a1+6d=3(a1+4d),解得a1=3d.又等差數(shù)列{an}是遞增數(shù)列,所以d>0,則a1<0,故選項(xiàng)A、選項(xiàng)B正確;因?yàn)镾n=d2n2+a1d2n=d2n27d2n,且-7d2d=72,所以當(dāng)n=3或4時(shí)Sn最小,故選項(xiàng)C錯(cuò)誤;令Sn=d2n27d2n>0,解得7.ABC解析:因?yàn)閍n+an+1=3n,n∈N*, ①所以an+1+an+2=3(n+1),n∈N*, ②所以②①得an+2an=3(n∈N*).又因?yàn)閍1=1,所以a2=2,所以a6=a4+3=a2+6=8,且奇數(shù)項(xiàng)和偶數(shù)項(xiàng)均為公差為3的等差數(shù)列,故A,B正確;對(duì)于C選項(xiàng),S20=(a1+a2)+(a3+a4)+…+(a19+a20)=3+9+…+57=10×(3+57)2=300,故C正確;對(duì)于D選項(xiàng),由an+2an=3(n∈N*)且a2a1≠3可知,a2na2n18.108解析:設(shè)等差數(shù)列{an}的公差為d,則d=2.因?yàn)閍1,a2,a3+1成等比數(shù)列,所以a1(a3+1)=a22,即a1(a1+5)=(a1+2)2,解得a1=4,所以S9=na1+n×(n-1)2×d=99.511解析:因?yàn)榈炔顢?shù)列{an},{bn}的前n項(xiàng)和分別是Sn,Tn所以a310.B解析:∵S2020>0,S2021<0,∴2020(a1+a2020)2=1010(a1010+a1011)>又2021×a1+a20212=2021a1011<0,∴a1010>0,a1011<0,可得a1010a1011<0,充分性成立.反之,若a1010<0,a1011>0,滿(mǎn)足a1010a1011<0,不能推出“S2020>0,S2021<0”,必要性不成立.故“S2020>0,S2021<0”是“a1010a1011<011.C解析:令S5=t,則由S5S10=13,得S10S5=2t,S10=3t.又由等差數(shù)列{an}的性質(zhì)得S5,S10S5,S15S10,S20S15成等差數(shù)列,所以有S10S5=2t,S15S10=3t,S20S15=4t,相加可得S20S5=9t,所以S20=10t12.ACD解析:依題意,S12=a1+a122·12=6(a6+a7)>0,于是得a6+a7>0,而a7<0,a6>a7>0,故選項(xiàng)B錯(cuò)誤;顯然有(a3+3d)+(a3+4d)>0,而a3=12,解得d>247,又a3+4d<0,解得d<3,因此得247<d<3,故選項(xiàng)A、選項(xiàng)C正確;數(shù)列{an}是首項(xiàng)為正數(shù),公差為負(fù)數(shù)的遞減數(shù)列,前6項(xiàng)都為正,從第7項(xiàng)起的各項(xiàng)都為負(fù),而S12>0,S13=a1+a132·13=13a7<0,于是得n≥13時(shí),S13.6解析:設(shè)等差數(shù)列{an}的公差為d,由等差中項(xiàng)的性質(zhì),得a1+a7=2a4=12,解得a4=6,所以a32+a42+a52=(6d)2+62+(6+d)2=2d2+108.當(dāng)d=14.84解析:依題意,冬至日晷長(zhǎng)為13.5尺,記為a1=13.5,芒種日晷長(zhǎng)為2.5尺,記為a12=2.5,因相鄰兩個(gè)節(jié)氣的日晷長(zhǎng)變化量相同,則從冬至日晷長(zhǎng)到芒種日晷長(zhǎng)的各數(shù)據(jù)依次排成一列得等差數(shù)列{an},n∈N*,n≤12,數(shù)列{an}的公差d=a12-因夏至與芒種相鄰,且夏至日晷長(zhǎng)最短,則夏至的日晷長(zhǎng)為a12+d=1.5,又大雪與冬至相鄰,且冬至日晷長(zhǎng)最長(zhǎng),則大雪的日晷長(zhǎng)為a1+d=12.5,顯然夏至到大雪的日晷長(zhǎng)依次排成一列是遞增等差數(shù)列,首項(xiàng)為1.5尺,末項(xiàng)為12.5尺,共12項(xiàng),所以一年中夏至到大雪的日晷長(zhǎng)的和為1.5+12.5215.解(1)由an+1=an+2得數(shù)列{an}是等差數(shù)列,其公差d=2,由S6=4a5得6a1+6×52d=4(a1+即6a1+30=4(a1+8),解得a1=1,所以an=1+2(n1)=2n1.(2)Sn=1+2n-12·n=n2,cosnπ所以bn=(1)n·n2,T2n=b1+b2+b3+b4+…+b2n1+b2n=12+2232+42…(2n1)2+(2n)2=1+2+3+4+…+2n1+2n=1+2n2·2n=2n216.BCD解析:由已知得Sn=a1n+n(n-1)dA選項(xiàng),S6=6a1+15d,S4=4a1+6d,S2=2a1+d,所以2S4S2=6a1+11d≠S6,A選項(xiàng)錯(cuò)誤;B選項(xiàng),3(S4S2)=6a1+15d=S6,B選項(xiàng)正確;C選項(xiàng),S2