板塊二數(shù)列微專題7等差數(shù)列與等比數(shù)列高考定位1.等差、等比數(shù)列的基本運(yùn)算和性質(zhì)的考查是高考熱點(diǎn),經(jīng)常以小題形式出現(xiàn);2.數(shù)列的通項(xiàng)也是高考熱點(diǎn),難度中檔以下.高考定位【

真題體驗(yàn)

】1.(2025·新高考Ⅱ卷)記Sn為等差數(shù)列{an}的前n項(xiàng)和.若S3=6,S5=-5,則S6= A.-20 B.-15 C.-10 D.-5√由S3=3a2=6,S5=5a3=-5,得a2=2,a3=-1,所以{an}的公差d=a3-a2=-3,所以a6=a3+3d=-10,所以S6=S5+a6=-5-10=-15.

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3.(2025·新高考Ⅰ卷)若一個等比數(shù)列的各項(xiàng)均為正數(shù),且前4項(xiàng)的和等于4,前8項(xiàng)的和等于68,則這個數(shù)列的公比等于

.

2

所以n≥2時,an=Sn-Sn-1=n2d2-(n-1)2d2=2d2n-d2,對n=1也適合,所以an=2d2n-d2,所以an+1-an=2d2(n+1)-d2-(2d2n-d2)=2d2(常數(shù)),所以數(shù)列{an}是等差數(shù)列.精準(zhǔn)強(qiáng)化練熱點(diǎn)一等差、等比數(shù)列的基本運(yùn)算熱點(diǎn)二等差、等比數(shù)列的性質(zhì)熱點(diǎn)三等差、等比數(shù)列的判定與證明熱點(diǎn)突破√

熱點(diǎn)一等差、等比數(shù)列的基本運(yùn)算例1√

等差、等比數(shù)列的基本量問題的求解(1)抓住基本量:首項(xiàng)a1、公差d或公比q.(2)熟悉一些結(jié)構(gòu)特征,如前n項(xiàng)和為Sn=an2+bn(a,b是常數(shù))形式的數(shù)列為等差數(shù)列,通項(xiàng)公式為an=p·qn-1(p,q≠0)形式的數(shù)列為等比數(shù)列.(3)由于等比數(shù)列的通項(xiàng)公式、前n項(xiàng)和公式中變量n在指數(shù)位置,所以常用兩式相除(即比值的方式)進(jìn)行相關(guān)計算.規(guī)律方法√(1)(2025·蚌埠二模)已知公差不為0的等差數(shù)列{an}的前n項(xiàng)和為Sn,若S15=5(a5+a7+ak),則正整數(shù)k的值為A.11 B.12 C.13 D.14

訓(xùn)練1√(2)(2025·北京卷)已知{an}是公差不為0的等差數(shù)列,a1=-2,若a3,a4,a6成等比數(shù)列,則a10=A.-20 B.-18 C.16 D.18

√(1)(2025·長春二模)已知{an}為正項(xiàng)等比數(shù)列,若lga2,lga2024是函數(shù)f(x)=3x2-12x+9的兩個零點(diǎn),則a1a2025=A.10 B.104 C.108

D.1012由題意可得lga2,lga2024為方程3x2-12x+9=0的兩個解,則lga2+lga2024=4,解得a2a2024=104,易知a1a2025=a2a2024=104.熱點(diǎn)二等差、等比數(shù)列的性質(zhì)例2√(2)(多選)設(shè)Sn是公差為d(d≠0)的無窮等差數(shù)列{an}的前n項(xiàng)和,則下列命題正確的是A.若d<0,則S1是數(shù)列{Sn}的最大項(xiàng)B.若數(shù)列{Sn}有最小項(xiàng),則d>0C.若數(shù)列{Sn}是遞減數(shù)列,則對任意的n∈N*,均有Sn<0D.若對任意的n∈N*,均有Sn>0,則數(shù)列{Sn}是遞增數(shù)列√

當(dāng)數(shù)列{Sn}有最小項(xiàng),即Sn有最小值時,Sn對應(yīng)的二次函數(shù)有最小值,對應(yīng)的函數(shù)圖象開口向上,d>0,B正確;對于C,取數(shù)列{an}為首項(xiàng)為1,公差為-2的等差數(shù)列,Sn=-n2+2n,Sn+1-Sn=-(n+1)2+2(n+1)-(-n2+2n)=-2n+1<0,即Sn+1<Sn恒成立,此時數(shù)列{Sn}是遞減數(shù)列,而S1=1>0,故C錯誤;對于D,若數(shù)列{Sn}是遞減數(shù)列,則an=Sn-Sn-1<0(n≥2),一定存在實(shí)數(shù)k,當(dāng)n>k時,之后所有項(xiàng)都為負(fù)數(shù),不能保證對任意n∈N*,均有Sn>0.故若對任意n∈N*,均有Sn>0,則數(shù)列{Sn}是遞增數(shù)列,故D正確.等差、等比數(shù)列性質(zhì)問題的求解(1)抓關(guān)系,抓住項(xiàng)與項(xiàng)之間的關(guān)系及項(xiàng)的序號之間的關(guān)系,從這些特點(diǎn)入手,選擇恰當(dāng)?shù)男再|(zhì)進(jìn)行求解.(2)用性質(zhì),數(shù)列是一種特殊的函數(shù),具有函數(shù)的一些性質(zhì),如單調(diào)性、周期性等,可利用函數(shù)的性質(zhì)解題.規(guī)律方法√(1)(2025·貴陽調(diào)研)已知等差數(shù)列{an}的前n項(xiàng)和為Sn,若S3=S9=6,則S12的值為A.0 B.3 C.6 D.12因?yàn)閧an}是等差數(shù)列,所以S3,S6-S3,S9-S6,S12-S9成等差數(shù)列,又S3=S9=6,所以6,S6-6,6-S6,S12-6成等差數(shù)列,則6+S12-6=S6-6+6-S6,則S12=0.訓(xùn)練2√

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(2025·佛山質(zhì)檢)已知數(shù)列{an}滿足a1=1,a2=3,an+2=3an+1-2an(n∈N*).(1)證明:數(shù)列{an+1-an}是等比數(shù)列;熱點(diǎn)三等差、等比數(shù)列的判定與證明例3∵an+2=3an+1-2an,∴an+2-an+1=2(an+1-an),∵a1=1,a2=3,∴{an+1-an}是以a2-a1=2為首項(xiàng),2為公比的等比數(shù)列.(2)求數(shù)列{an}的通項(xiàng)公式;由(1)得an+1-an=2n(n∈N*),∴an=(an-an-1)+(an-1-an-2)+…+(a2-a1)+a1=2n-1+2n-2+…+2+1=2n-1(n≥2),又a1=1符合上式,∴an=2n-1(n∈N*).

④-③,得nbn+2-2nbn+1+nbn=0,即bn+2-2bn+1+bn=0,∴bn+2-bn+1=bn+1-bn(n∈N*),∴{bn}是等差數(shù)列.1.判斷等差、等比數(shù)列的常用方法規(guī)律方法

等差數(shù)列等比數(shù)列定義法an+1-an=d通項(xiàng)法an=a1+(n-1)dan=a1·qn-1中項(xiàng)法2an=an-1+an+1(n≥2)前n項(xiàng)和法Sn=an2+bn(a,b為常數(shù))Sn=kqn-k(k≠0,q≠0,1)2.證明數(shù)列為等差(比)數(shù)列一般使用定義法.規(guī)律方法(2025·山東名校調(diào)研)已知數(shù)列{an},{bn}是公比不相等的兩個等比數(shù)列,令cn=an+bn.(1)證明:數(shù)列{cn}不是等比數(shù)列.訓(xùn)練3

(2)若an=2n,bn=3n,是否存在常數(shù)k,使得數(shù)列{cn+1+kcn}為等比數(shù)列?若存在,求出k的值;若不存在,請說明理由.假設(shè)存在常數(shù)k,使得數(shù)列{cn+1+kcn}為等比數(shù)列,則有(cn+1+kcn)2=(cn+2+kcn+1)(cn+kcn-1),n≥2,將cn=2n+3n代入上式,得[2n+1+3n+1+k(2n+3n)]2=[2n+2+3n+2+k(2n+1+3n+1)]·[2n+3n+k(2n-1+3n-1)],即[(2+k)2n+(3+k)3n]2=[(2+k)2n+1+(3+k)3n+1]·[(2+k)2n-1+(3+k)3n-1],整理得12(2+k)(3+k)=13(2+k)(3+k),解得k=-2或k=-3.當(dāng)k=-2時,cn+1+kcn=2n+1+3n+1+(-2)·(2n+3n)=3n,此時數(shù)列{cn+1+kcn}為等比數(shù)列;當(dāng)k=-3時,cn+1+kcn=2n+1+3n+1+(-3)·(2n+3n)=-2n,數(shù)列{cn+1+kcn}為等比數(shù)列.所以存在常數(shù)k=-2或k=-3,使得數(shù)列{cn+1+kcn}為等比數(shù)列.【精準(zhǔn)強(qiáng)化練】√1.(2025·濰坊模擬)已知等差數(shù)列{an}的前n項(xiàng)和為Sn,若a1=2,a3+a9=24,則S6= A.12 B.14 C.42 D.84

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√3.(2025·天津紅橋區(qū)模擬)等比數(shù)列{an}的前n項(xiàng)和為Sn,且a1+a4=4,a2+a5=8,則S6= A.24 B.28 C.36 D.48

√4.等比數(shù)列{an}的前n項(xiàng)和為Sn,若S10=10,S20=30,則S40= A.60 B.70 C.80 D.150由題意得q≠-1,且{an}是等比數(shù)列,所以S10,S20-S10,S30-S20,S40-S30成等比數(shù)列,又因?yàn)镾10=10,S20=30,S20-S10=20,則S30-S20=40,S40-S30=80,所以S30=70,S40=150.√

√6.在各項(xiàng)均為正數(shù)的等比數(shù)列{an}中,已知a2>1,其前n項(xiàng)積為Tn,且T20=T10,則Tn取得最大值時,n的值為 A.15 B.16 C.29 D.30

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A.41米 B.40.5米C.39.5米 D.38.7米

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12.(2025·渭南模擬)已知等差數(shù)列{an}的前n項(xiàng)和為Sn,由S13=6.則3a9-2a10=

.

14.(2025·宜荊荊恩調(diào)研)已知數(shù)列{an}有30項(xiàng),a1=2,且對任意n∈{2,3,…,30},都存在i∈{1,2,…,n-1},使得an=ai+3. (1)a5=

;(寫出所有可能的取值)

當(dāng)n=2時,a2=a1+3=5,當(dāng)n=3時,a3=a1+3=5或a3=a2+3=8,當(dāng)n=4時,a4=a1+3=5或a4=a2+3=8或a4=a3+3時有a4=8或a4=11,當(dāng)n=5時,a5=a1+3=5或a5=a2+3=8或a5=a3+3時有a5=8或a5=11或a5=a4+3時有a5=8或a5=11或