2023版高考数学一轮总复习 10年高考真题分类题组 6.docx

1、6.2等差数列考点一等差数列及其前n项和1.(2015课标文,5,5分)设Sn是等差数列an的前n项和.若a1+a3+a5=3,则S5=()A.5B.7C.9D.11答案Aan为等差数列,a1+a5=2a3,得3a3=3,则a3=1,S5=5(a1+a5)2=5a3=5,故选A.2.(2015课标文,7,5分)已知an是公差为1的等差数列,Sn为an的前n项和.若S8=4S4,则a10=()A.172B.192C.10D.12答案B由S8=4S4得8a1+8721=44a1+4321,解得a1=12,a10=a1+9d=192,故选B.评析本题主要考查等差数列的前n项和,计算准确是解题关键,属

2、容易题.3.(2015浙江理,3,5分)已知an是等差数列,公差d不为零,前n项和是Sn.若a3,a4,a8成等比数列,则()A.a1d0,dS40B.a1d0,dS40,dS40D.a1d0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,则a1d=-53d20,又S4=4a1+6d=-23d,dS4=-23d21.评析本题考查等差、等比数列的基础知识,考查运算求解能力.11.(2016天津,18,13分)已知an是各项均为正数的等差数列,公差为d.对任意的nN*,bn是an和an+1的等比中项.(1)设cn=b

3、n+12-bn2,nN*,求证:数列cn是等差数列;(2)设a1=d,Tn=k=12n(-1)kbk2,nN*,求证:k=1n1Tk12d2.证明(1)由题意得bn2=anan+1,有cn=bn+12-bn2=an+1an+2-anan+1=2dan+1,因此cn+1-cn=2d(an+2-an+1)=2d2,所以cn是等差数列.(2)Tn=(-b12+b22)+(-b32+b42)+(-b2n-12+b2n2)=2d(a2+a4+a2n)=2dn(a2+a2n)2=2d2n(n+1).所以k=1n1Tk=12d2k=1n1k(k+1)=12d2k=1n1k-1k+1=12d21-1n+10,a7+a100,即a80.又a8+a9=a7+a100,a90,当n=8时,an的前n项和最大.