课时规范练28等差数列及其前n项和基础巩固组1.由a1=1,d=3确定的等差数列{an},当an=298时,序号n等于()A.99B.100C.96D.1012.(2018湖南长郡中学仿真,6)已知等差数列{an}满足an+1+an=4n,则a1=()A.-1B.1C.2D.33.(2018河南商丘二模,3)已知等差数列{an}的公差为d,且a8+a9+a10=24,则a1·d的最大值为()A.B.C.2D.44.在等差数列{an}中,a3+a6=11,a5+a8=39,则公差d为()A.-14B.-7C.7D.145.已知等差数列{an}的前n项和为Sn,且3a3=a6+4,若S5<10,则a2的取值范围是()A.(-∞,2)B.(-∞,0)C.(1,+∞)D.(0,2)6.已知Sn是等差数列{an}的前n项和,则2(a1+a3+a5)+3(a8+a10)=36,则S11=()A.66B.55C.44D.337.(2018湖南衡阳一模,15)已知数列{an}前n项和为Sn,若Sn=2an-2n,则Sn=.8.设数列{an}{bn}都是等差数列,若a1+b1=7,a3+b3=21,则a5+b5=.9.若数列{an}的前n项和为Sn,且满足an+2SnSn-1=0(n≥2),a1=.(1)求证:成等差数列;(2)求数列{an}的通项公式.10.设数列{an}的前n项和为Sn,且Sn=2n-1.数列{bn}满足b1=2,bn+1-2bn=8an.(1)求数列{an}的通项公式;(2)证明:数列为等差数列,并求{bn}的通项公式.综合提升组11.(2018河北衡水中学考前押题二,10)已知数列a1=1,a2=2,且an+2-an=2-2(-1)n,n∈N+,则S2017的值为()A.2016×1010-1B.1009×2017C.2017×1010-1D.1009×201612.若数列{an}满足:a1=19,an+1=an-3(n∈N+),则数列{an}的前n项和数值最大时,n的值为()A.6B.7C.8D.913.已知等差数列{an}的前n项和为Sn,若Sm-1=-4,Sm=0,Sm+2=14(m≥2,且m∈N+),则m的值为.14.已知公差大于零的等差数列{an}的前n项和为Sn,且满足a3·a4=117,a2+a5=22.(1)求通项公式an;(2)求Sn的最小值;(3)若数列{bn}是等差数列,且bn=,求非零常数c.创新应用组15.(2018湖南长郡中学仿真,15)若数列{an}是正项数列,且+…+=n2+3n,则+…+=.16.等差数列{an}的前n项和为Sn,已知S10=0,S15=25,则nSn的最小值为多少?1课时规范练28等差数列及其前n项和1.B根据等差数列通项公式an=a1+(n-1)d,有298=1+(n-1)×3,解得n=100,故选B.2.B由题意,当n分别取1,2时,a1+a2=4,a3+a2=8,解得公差d=2,故a1=1.故选B.3.C∵a8+a9+a10=24,∴a9=8,即a1+8d=8,∴a1=8-8d,a1·d=(8-8d)d=-8d-2+2≤2,当d=时,a1·d的最大值为2,故选C.4.C∵a3+a6=11,a5+a8=39,则4d=28,解得d=7.故选C.5.A设公差为d,由3a3=a6+4得3(a2+d)=a2+4d+4,即d=2a2-4,由S5<10得,=5(3a2-4)<10,解得a2<2,故选A.6.D由等差数列的性质可得2(a1+a3+a5)+3(a8+a10)=6a3+6a9=36,即a1+a11=6.则S11==11×3=33.故选D.7.n·2n∵Sn=2an-2n=2(Sn-Sn-1)-2n,整理得Sn-2Sn-1=2n,等式两边同时除以2n,则=1.又S1=2a1-2=a1,可得a1=S1=2,∴数列是以1为首项,公差为1的等差数列,∴=n,∴Sn=n·2n.8.35∵数列{an},{bn}都是等差数列,设数列{an}的公差为d1,数列{bn}的公差为d2,∴a3+b3=a1+b1+2(d1+d2)=21,而a1+b1=7,可得2(d1+d2)=21-7=14.∴a5+b5=a3+b3+2(d1+d2)=21+14=35.9.(1)证明当n≥2时,由an+2SnSn-1=0,得Sn-Sn-1=-2SnSn-1,∴=2.又=2,故是首项为2,公差为2的等差数列.(2)解由(1)可得=2n,∴Sn=.当n≥2时,an=Sn-Sn-1==-.当n=1时,a1=不适合上式.故an=10.(1)解当n=1时,a1=S1=21-1=1;当n≥2时,an=Sn-Sn-1=(2n-1)-(2n-1-1)=2n-1.∵a1=1适合通项公式an=2n-1,∴an=2n-1.(2)证明∵bn+1-2bn=8an,∴bn+1-2bn=2n+2,即=2.又=1,∴是首项为1,公差为2的等差数列.∴=1+2(n-1)=2n-1.∴bn=(2n-1)×2n.11.C由题意,当n为奇数时,an+2-an=4,数列{a2n-1}是首项为1,公差为4的等差数列,当n为偶数时,an+2-an=0,数列{a2n-1}是首项为2,公差为0的等差数列,S2017=(a1+a3+…+a2017)+(a2+a4+…+a2016)=1009+×1009×1008×4+1008×2=2017×1010-1,故选C.12.B∵a1=19,an+1-an=-3,∴数列{an}是以19为首项,-3为公差的等差数列.∴an=19+(n-1)×(-3)=22-3n.设数列{an}的前k项和数值最大,则有k∈N+.∴2∴≤k≤.∵k∈N+,∴k=7.∴满足条件的n的值为7.13.5∵Sm-1=-4,Sm=0,Sm+2=14,∴am=Sm-Sm-1=4,am+1+am+2=Sm+2-Sm=14.设数列{an}的公差为d,则2am+3d=14,∴d=2.∵Sm=×m=0,∴a1=-am=-4.∴am=a1+(m-1)d=-4+2(m-1)=4,∴m=5.14.解(1)∵数列{an}为等差数列,∴a3+a4=a2+a5=22.又a3·a4=117,∴a3,a4是方程x2-22x+117=0的两实根.又公差d>0,∴a3
时,f'(n)>0,当0
