2013-2022年數(shù)學(xué)高考真題專題練習(xí)--等差數(shù)列

1、2013-2022年數(shù)學(xué)高考真題專題練習(xí)7.2等差數(shù)列考點(diǎn)一等差數(shù)列及其前n項(xiàng)和1.(2015課標(biāo)文,5,5分)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Aan為等差數(shù)列,a1+a5=2a3,得3a3=3,則a3=1,S5=5(a1+a5)2=5a3=5,故選A.2.(2015課標(biāo)文,7,5分)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.172B.192C.10D.12答案B由S8=4S4得8a1+8×72×1=4×4a1+4×32×1,解得a1=

2、12,a10=a1+9d=192,故選B.評析本題主要考查等差數(shù)列的前n項(xiàng)和,計(jì)算準(zhǔn)確是解題關(guān)鍵,屬容易題.3.(2015浙江理,3,5分)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.a1d>0,dS4>0B.a1d<0,dS4<0C.a1d>0,dS4<0D.a1d<0,dS4>0答案B由a42=a3a8,得(a1+2d)(a1+7d)=(a1+3d)2,整理得d(5d+3a1)=0,又d0,a1=-53d,則a1d=-53d2<0,又S4=4a1+6d=-23d,dS4=-23d2<0

3、,故選B.4.(2013課標(biāo)理,7,5分)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.6答案C解法一:Sm-1=-2,Sm=0,Sm+1=3,am=Sm-Sm-1=2,am+1=Sm+1-Sm=3,公差d=am+1-am=1,由Sn=na1+n(n1)2d=na1+n(n1)2,得ma1+m(m1)2=0,(m1)a1+(m1)(m2)2=2.由得a1=1m2,代入可得m=5.解法二:數(shù)列an為等差數(shù)列,且前n項(xiàng)和為Sn,數(shù)列Snn也為等差數(shù)列.Sm1m1+Sm+1m+1=2Smm,即2m1+3m+1=0,解得m=5.經(jīng)檢驗(yàn)為原方程

4、的解.故選C.5.(2016課標(biāo),3,5分)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C設(shè)an的公差為d,由等差數(shù)列前n項(xiàng)和公式及通項(xiàng)公式,得S9=9a1+9×82d=27,a10=a1+9d=8,解得a1=1,d=1,an=a1+(n-1)d=n-2,a100=100-2=98.故選C.思路分析用a1,d表示S9,a10,列方程組求出a1,d,從而可求得a100.6.(2014天津理,11,5分)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. 答案-12解析

5、S1=a1,S2=2a1-1,S4=4a1-6.故(2a1-1)2=a1×(4a1-6),解得a1=-12.7.(2013課標(biāo)理,16,5分)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. 答案-49解析由Sn=na1+n(n1)2d得10a1+45d=0,15a1+105d=25,解得a1=-3,d=23,則Sn=-3n+n(n1)2·23=13(n2-10n),所以nSn=13(n3-10n2),令f(x)=13(x3-10x2),則 f '(x)=x2-203x=xx203,當(dāng)x1,203時, f(x)遞減,當(dāng)x20

6、3,+時, f(x)遞增,又6<203<7, f(6)=-48,f(7)=-49,所以nSn的最小值為-49.評析本題考查了數(shù)列與函數(shù)的應(yīng)用,考查了數(shù)列的基本運(yùn)算,利用導(dǎo)數(shù)求最值.本題易忽略n的取值范圍.8.(2013課標(biāo)文,17,12分)已知等差數(shù)列an的公差不為零,a1=25,且a1,a11,a13成等比數(shù)列.(1)求an的通項(xiàng)公式;(2)求a1+a4+a7+a3n-2.解析(1)設(shè)an的公差為d.由題意,得a112=a1a13,即(a1+10d)2=a1(a1+12d).于是d(2a1+25d)=0.又a1=25,所以d=0(舍去)或d=-2.故an=-2n+27.(2)令S

7、n=a1+a4+a7+a3n-2.由(1)知a3n-2=-6n+31,故a3n-2是首項(xiàng)為25,公差為-6的等差數(shù)列.從而Sn=n2(a1+a3n-2)=n2(-6n+56)=-3n2+28n.9.(2012湖北,理18,文20,12分)已知等差數(shù)列an前三項(xiàng)的和為-3,前三項(xiàng)的積為8.(1)求等差數(shù)列an的通項(xiàng)公式;(2)若a2,a3,a1成等比數(shù)列,求數(shù)列|an|的前n項(xiàng)和.解析(1)設(shè)等差數(shù)列an的公差為d,則a2=a1+d,a3=a1+2d,由題意得3a1+3d=3,a1(a1+d)(a1+2d)=8,解得a1=2,d=3或a1=4,d=3.所以由等差數(shù)列通項(xiàng)公式可得an=2-3(n-

8、1)=-3n+5或an=-4+3(n-1)=3n-7.故an=-3n+5或an=3n-7.(2)當(dāng)an=-3n+5時,a2,a3,a1分別為-1,-4,2,不成等比數(shù)列;當(dāng)an=3n-7時,a2,a3,a1分別為-1,2,-4,成等比數(shù)列,滿足條件.故|an|=|3n-7|=3n+7,n=1,2,3n7,n3.記數(shù)列|an|的前n項(xiàng)和為Sn.當(dāng)n=1時,S1=|a1|=4;當(dāng)n=2時,S2=|a1|+|a2|=5;當(dāng)n3時,Sn=S2+|a3|+|a4|+|an|=5+(3×3-7)+(3×4-7)+(3n-7)=5+(n2)2+(3n7)2=32n2-112n+10.當(dāng)n

9、=2時,滿足此式,綜上,Sn=4,n=1,32n2112n+10,n>1.評析本題考查等差、等比數(shù)列的基礎(chǔ)知識,考查運(yùn)算求解能力.10.(2014大綱全國文,17,10分)數(shù)列an滿足a1=1,a2=2,an+2=2an+1-an+2.(1)設(shè)bn=an+1-an,證明bn是等差數(shù)列;(2)求an的通項(xiàng)公式.解析(1)證明:由an+2=2an+1-an+2得,an+2-an+1=an+1-an+2,即bn+1=bn+2.又b1=a2-a1=1,所以bn是首項(xiàng)為1,公差為2的等差數(shù)列.(5分)(2)由(1)得bn=1+2(n-1),即an+1-an=2n-1.(8分)于是k=1n(ak+1

10、ak)=k=1n(2k1),所以an+1-a1=n2,即an+1=n2+a1.又a1=1,所以an的通項(xiàng)公式為an=n2-2n+2.(10分)評析本題著重考查等差數(shù)列的定義、前n項(xiàng)和公式及“累加法”求數(shù)列的通項(xiàng)等基礎(chǔ)知識,同時考查運(yùn)算變形的能力.11.(2014課標(biāo)理,17,12分)已知數(shù)列an的前n項(xiàng)和為Sn,a1=1,an0,anan+1=Sn-1,其中為常數(shù),(1)證明:an+2-an=;(2)是否存在,使得an為等差數(shù)列?并說明理由.解析(1)證明:由題設(shè)anan+1=Sn-1,知an+1an+2=Sn+1-1.兩式相減得,an+1(an+2-an)=an+1.由于an+10,所以an

11、+2-an=.(2)存在.由a1=1,a1a2=a1-1,可得a2=-1,由(1)知,a3=+1.令2a2=a1+a3,解得=4.故an+2-an=4,由此可得,a2n-1是首項(xiàng)為1,公差為4的等差數(shù)列,a2n-1=1+(n-1)·4=4n-3;a2n是首項(xiàng)為3,公差為4的等差數(shù)列,a2n=3+(n-1)·4=4n-1.所以an=2n-1,an+1-an=2.因此存在=4,使得an為等差數(shù)列.評析本題主要考查an與Sn的關(guān)系及等差數(shù)列的定義,考查學(xué)生的邏輯思維能力及分析解決問題的能力.考點(diǎn)二等差數(shù)列的性質(zhì)1.(2015廣東理,10,5分)在等差數(shù)列an中,若a3+a4+a5+a6+a7=25,則a2+a8=. 答案10解析利用等差數(shù)列的性質(zhì)可得a3+a7=a4+a6=2a5,從而a3+a4+a5+a6+a7=5a5=25,故a5=5,所以a2+a8=2a5=10.2.(2015陜西文,13,5分)中位數(shù)為1 010的一組數(shù)構(gòu)成等差數(shù)列,其末項(xiàng)為2 015,則該數(shù)列的首項(xiàng)為. 答案5解析設(shè)該數(shù)列的首項(xiàng)為a1,根據(jù)等差數(shù)列的性質(zhì)可得a1+2 015=2