2025版高考數(shù)學(xué)一輪復(fù)習(xí)核心考點(diǎn)精準(zhǔn)研析8.2等差數(shù)列文含解析北師大版

PAGE11-等差數(shù)列核心考點(diǎn)·精準(zhǔn)研析考點(diǎn)一等差數(shù)列的基本運(yùn)算

1.在等差數(shù)列{an}中,a1+a5=8,a4=7,則a5= ()A.11 B.10 C.7 D.32.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.172 B.192 C.103.(2024·沈陽(yáng)模擬)在等差數(shù)列{an}中,若Sn為前n項(xiàng)和,2a7=a8+5,則S11的值是()A.55 B.11 C.50 D.604.(2024·全國(guó)卷Ⅰ)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-5 B.an=3n-10C.Sn=2n2-8n D.Sn=12n25.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=________.

【解析】1.選B.設(shè)等差數(shù)列{an}的公差為d,則有2a1+4d=8,2.選B.由公差為1得S8=8a1+8×(8-1)2×1=8a1+28,S4=4a1+6.因?yàn)镾8=4S4,所以8a1+28=4(4a1+6),解得a1=12,所以a3.選A.設(shè)等差數(shù)列{an}的公差為d,由題意可得2(a1+6d)=a1+7d+5,得a1+5d=5,則S11=11a1+11×102d=11(a14.選A.設(shè)等差數(shù)列{an}的公差為d,由題知,S4=4a1+5.由Sm-1=-2,Sm=0,Sm+1=3得am=Sm-Sm-1=2,am+1=Sm+1-Sm=3,所以等差數(shù)列的公差d=am+1-am=3-2=1,由a得a1+答案:5第3題中若將條件“2a7=a8+5”改為“a9=12a12+6”,其他條件不變,則數(shù)列{an}的前11項(xiàng)和S11【解析】S11=11(a1+設(shè)公差為d,由a9=12a12得a6+3d=12(a6解得a6=12,所以S11=11×12=132.答案:132【接著探究】已知等差數(shù)列{an}的前n項(xiàng)和為Sn,且a9=12a12+6,a2=4,則數(shù)列1Sn的前2A.20202021 B.20212【解析】選D.設(shè)等差數(shù)列{an}的公差為d,由a9=12a12+6及等差數(shù)列的通項(xiàng)公式得a1+5d=12,又a2=4,所以a1所以Sn=n2+n,所以1Sn=1n(n所以1S1+1S2+…+1S2021=1-12等差數(shù)列運(yùn)算問(wèn)題的通性方法1.等差數(shù)列運(yùn)算問(wèn)題的一般求法是設(shè)出首項(xiàng)a1和公差d,然后由通項(xiàng)公式或前n項(xiàng)和公式轉(zhuǎn)化為方程(組)求解.2.等差數(shù)列的通項(xiàng)公式及前n項(xiàng)和公式,共涉及五個(gè)量a1,an,d,n,Sn,知其中三個(gè)就能求另外兩個(gè),體現(xiàn)了用方程的思想解決問(wèn)題.【秒殺絕技】1.應(yīng)用性質(zhì)解T1由等差數(shù)列的性質(zhì)得a1+a5=2a3=8,所以a3=4,故d=a4-a3=3.所以a5=a4+d=10.2.應(yīng)用變形公式解T3設(shè)等差數(shù)列{an}的公差為d,由2a7=a8+5,得2(a6+d)=a6+2d+5,得a6=5,所以S11=11a6=55.3.應(yīng)用解除法解T4對(duì)于B,a5=5,S4=4(-7+2)2=-10對(duì)于C,S4=0,a5=S5-S4=2×52-8×5-0=10≠5,解除C.對(duì)于D,S4=0,a5=S5-S4=12×52-2×5-0=2.5≠5,解除D,故選考點(diǎn)二等差數(shù)列的判定與證明

【典例】1.(2024·貴陽(yáng)模擬)已知數(shù)列{an}滿(mǎn)意a1=1,且nan+1-(n+1)an=2n2+2n. 世紀(jì)金榜導(dǎo)學(xué)號(hào)(1)求a2,a3;(2)證明數(shù)列ann是等差數(shù)列,并求{an}【解題導(dǎo)思】序號(hào)題目拆解(1)①a1=1,且nan+1-(n+1)an=2n2+2n代入n=1得a2②由a2及nan+1-(n+1)an=2n2+2n代入n=2得a3(2)①nan+1-(n+1)an=2n2+2nnan+1-(n+1)an=2n2+2n變形為nan+1-(n+1)an=2n(n+1),結(jié)合所求結(jié)論,式子兩邊同除以n(n+1),證明數(shù)列an②數(shù)列an依據(jù)數(shù)列ann是等差數(shù)列寫(xiě)出{an【解析】(1)由已知,得a2-2a1=4,則a2=2a1+4,又a1=1,所以a2=6.由2a3-3a2=12,得2a3=12+3a2,所以a3=15.(2)由已知nan+1-(n+1)an=2n2+2n,得na即an+1n+1-an則ann=1+2(n-1)=2n-1,所以an=2n2.數(shù)列{an}滿(mǎn)意a1=1,an+1=(n2+n-λ)an(n∈N*),(1)當(dāng)a2=-1時(shí),求λ的值及a3的值;(2)是否存在λ,使數(shù)列{an}為等差數(shù)列?若存在,求出其通項(xiàng)公式,若不存在,說(shuō)明理由.【解題導(dǎo)思】序號(hào)聯(lián)想解題(1)看到an+1=(n2+n-λ)an,想到數(shù)列的遞推公式(2)看到an+1=(n2+n-λ)an,結(jié)合(1)想到若數(shù)列{an}為等差數(shù)列,可求λ,結(jié)合等差數(shù)列的定義推斷【解析】(1)因?yàn)閍n+1=(n2+n-λ)an,a1=1,a2=-1,所以-1=(2-λ)×1,解得λ=3.所以a3=(22+2-3)×(-1)=-3.(2)不存在λ,使數(shù)列{an}為等差數(shù)列,說(shuō)明如下:因?yàn)閍1=1,an+1=(n2+n-λ)an(n∈N*).所以,a2=2-λ,a3=(6-λ)(2-λ),a4=(12-λ)(6-λ)(2-λ),若存在實(shí)數(shù)λ,使數(shù)列{an}為等差數(shù)列.則a1+a3=2a2,即1+(6-λ)(2-λ)=2(2-λ),解得:λ=3.此時(shí)a2=2-λ=2-3=-1,a3=(6-λ)(2-λ)=-3,a4=(12-λ)(6-λ)(2-λ)=-27,a2-a1=-1-1=-2,而a4-a3=-24.所以,數(shù)列{an}不是等差數(shù)列,即不存在λ使數(shù)列{an}為等差數(shù)列.1.推斷數(shù)列{an}是等差數(shù)列的常用方法(1)定義法:對(duì)隨意n∈N*,an+1-an是同一常數(shù).(2)等差中項(xiàng)法:對(duì)隨意n≥2,n∈N*,滿(mǎn)意2an=an+1+an-1.(3)通項(xiàng)公式法:對(duì)隨意n∈N*,都滿(mǎn)意an=pn+q(p,q為常數(shù)).(4)前n項(xiàng)和公式法:對(duì)隨意n∈N*,都滿(mǎn)意Sn=An2+Bn(A,B為常數(shù)).說(shuō)明:證明數(shù)列{an}是等差數(shù)列的最終方法只能用定義法和等差中項(xiàng)法.2.證明某數(shù)列不是等差數(shù)列若證明某數(shù)列不是等差數(shù)列,則只要證明存在連續(xù)三項(xiàng)不成等差數(shù)列即可.(2024·齊齊哈爾模擬)已知數(shù)列{an}是等差數(shù)列,且a1,a2(a1<a2)分別為方程x2-6x+5=0的兩個(gè)根.(1)求數(shù)列{an}的前n項(xiàng)和Sn;(2)在(1)中,設(shè)bn=Snn+c,求證:當(dāng)c=-12時(shí),數(shù)列【解析】(1)因?yàn)閍1,a2(a1<a2)分別為方程x2-6x+5=0的兩個(gè)根,所以a1=1,a2=5,所以等差數(shù)列{an}的公差為4,所以Sn=n·1+n(n-1(2)當(dāng)c=-12時(shí),bn=Snn因?yàn)閎n+1-bn=2(n+1)-2n=2,b1=2,所以{bn}是首項(xiàng)為2,公差為2的等差數(shù)列.考點(diǎn)三等差數(shù)列性質(zhì)及其應(yīng)用

命題精解讀1.考什么:(1)等差數(shù)列性質(zhì).(2)等差數(shù)列前n項(xiàng)和的最值2.怎么考:等差數(shù)列性質(zhì)作為考查等差數(shù)列運(yùn)算學(xué)問(wèn)的最佳載體,因其考查學(xué)問(wèn)點(diǎn)較多成為高考命題的熱點(diǎn)3.新趨勢(shì):解題過(guò)程中經(jīng)常滲透數(shù)學(xué)運(yùn)算的核心素養(yǎng).學(xué)霸好方法1.等差數(shù)列常用性質(zhì)和結(jié)論的運(yùn)用2.求等差數(shù)列前n項(xiàng)和Sn最值的兩種方法(1)函數(shù)法(2)通項(xiàng)變號(hào)法3.交匯問(wèn)題數(shù)列與不等式結(jié)合考查分類(lèi)探討思想、數(shù)列與函數(shù)結(jié)合考查數(shù)形結(jié)合思想與等差數(shù)列項(xiàng)的性質(zhì)有關(guān)的運(yùn)算【典例】1.(2024·武漢模擬)在等差數(shù)列{an}中,前n項(xiàng)和Sn滿(mǎn)意S7-S2=45,則a5= ()A.7 B.9 C.14 D.18【解析】選B.因?yàn)樵诘炔顢?shù)列{an}中,S7-S2=45,所以a3+a4+a5+a6+a7=5a5=45,所以a5=9.【一題多解】選B.設(shè)等差數(shù)列{an}的公差為d,因?yàn)樵诘炔顢?shù)列{an}中,S7-S2=45,所以7a1+7×62d-(2a1+d)=45,整理得a12.(2024·太原模擬)在等差數(shù)列{an}中,2(a1+a3+a5)+3(a8+a10)=36,則a6= 世紀(jì)金榜導(dǎo)學(xué)號(hào)()A.8 B.6 C.4 D.3【解析】選D.由等差數(shù)列的性質(zhì)可知2(a1+a3+a5)+3(a8+a10)=2×3a3+3×2a9=6×2a6=36,得a6=3.在等差數(shù)列中涉及兩項(xiàng)和時(shí),應(yīng)用哪些性質(zhì)能夠幫助我們快速解題?提示:在等差數(shù)列中涉及兩項(xiàng)和時(shí),肯定要留意其項(xiàng)數(shù)和的關(guān)系,假如和相等,則兩項(xiàng)的和也對(duì)應(yīng)相等.等差數(shù)列和的性質(zhì)【典例】1.一個(gè)正項(xiàng)等差數(shù)列前n項(xiàng)的和為3,前3n項(xiàng)的和為21,則前2n項(xiàng)的和為 ()A.18 B.12 C.10 D.6【解析】選C.設(shè)此數(shù)列為{an},因?yàn)閧an}是等差數(shù)列,所以Sn,S2n-Sn,S3n-S2n成等差數(shù)列,即2(S2n-Sn)=Sn+(S3n-S23+21-S2n,解得2.等差數(shù)列{an},{bn}的前n項(xiàng)和分別為Sn,Tn,若anbn=2n3n+1,【解析】由anbn=2an2bn=a1+a2n-1b1+答案:11在等差數(shù)列中Sn,S2n-Sn,S3n-S2n,…成等差數(shù)列嗎?提示:在等差數(shù)列中Sn,S2n-Sn,S3n-S2n,…仍成等差數(shù)列.等差數(shù)列和的最值問(wèn)題【典例】等差數(shù)列{an}中,a1>0,S5=S12,則當(dāng)Sn有最大值時(shí),n的值為_(kāi)_______. 世紀(jì)金榜導(dǎo)學(xué)號(hào)

【解析】設(shè)等差數(shù)列{an}的公差為d,由S5=S12得5a1+10d=12a1+66d,所以d=-18a1<0.設(shè)此數(shù)列的前n項(xiàng)和最大,則即an=a1+(n-1)·-18答案:8或9【一題多解】方法一:由S5=S12,得a6+a7+a8+a9+a10+a11+a12=0,即7a9=0,a9=0,由a1>0可知d<0,故當(dāng)n=8或n=9時(shí)Sn最大.方法二:由S5=S12,可得a1=-8d,所以Sn=-8dn+n(n-1)2d=d2(n-17答案:8或9在涉及等差數(shù)列前n項(xiàng)和的問(wèn)題時(shí),你能總結(jié)出求該數(shù)列的前n項(xiàng)和Sn的最大(小)值的方法嗎?提示:求等差數(shù)列前n項(xiàng)和的最值的方法(1)二次函數(shù)法:用求二次函數(shù)最值的方法(配方法)求其前n項(xiàng)和的最值,但要留意n∈N*.(2)圖像法:利用二次函數(shù)圖像的對(duì)稱(chēng)性來(lái)確定n的值,使Sn取得最值.(3)項(xiàng)的符號(hào)法:當(dāng)a1>0,d<0時(shí),滿(mǎn)意an≥0,an+1≤01.(2024·濟(jì)南模擬)等差數(shù)列{an}的前n項(xiàng)和為Sn,若a1+a3+a5+a7+a9=20,則S9=()A.27 B.36 C.45 D.54【解析】選B.依題意a1+a3+a5+a7+a9=5a5=20,a5=4,所以S9=a1+a922.已知數(shù)列{an}的通項(xiàng)公式an=26-2n,若使此數(shù)列的前n項(xiàng)和Sn最大,則n的值為()A.12 B.13 C.12或13 D.14【解析】選C.因?yàn)閍n是遞減數(shù)列,a12=2,a13=0,所以S12=S13最大,所以n的值為12或【變式備選】設(shè)等差數(shù)列{an}的前n項(xiàng)和為Sn,若S6>S7>S5,則滿(mǎn)意SnSn+1<0的正整數(shù)n的值為()A.10 B.11 C.12 D.13【解析】選C.由S6>S7>S5,得S7=S6+a7<S6,S7=S5+a6+a7>S5,所以a7<0,a6+a7>0,所以S13=13(a1+a13)2=13a7<0,S12=12(a即滿(mǎn)意SnSn+1<0的正整數(shù)n的值為12.3.(2024·長(zhǎng)沙一中模擬)等差數(shù)列{an}的前n項(xiàng)和為Sn,已知S3=12,S6=51,則S9的值等于 ()A.66 B.90 C.117 D.127【解析】選C.等差數(shù)列{an}的前n項(xiàng)和為Sn,由題意可得S3,S6-S3,S9-S6成等差數(shù)列,故2(S6-S3)=S3+(S9-S6),代入數(shù)據(jù)可得2×(51-12)=12+(S9-51),解得S9=117.1.在等差數(shù)列{an}中,已知a3+a8=10,則3a5+a7= ()A.10 B.18 C.20 D.28【解析】選C.因?yàn)閍3+a8=10,所以由等差數(shù)列的性