第四章数列4.2等差数列4.2.1等差数列的概念第1课时等差数列的概念及通项公式课后篇巩固提升基础达标练1.已知等差数列{an}的首项a1=2,公差d=3,则数列{an}的通项公式为()A.an=3n-1B.an=2n+1C.an=2n+3D.an=3n+2解析an=a1+(n-1)d=2+(n-1)·3=3n-1.答案A2.若△ABC的三个内角A,B,C成等差数列,则cos(A+C)=()A.12B.√32C.-12D.-√32解析因为A,B,C成等差数列,所以A+C=2B.又因为A+B+C=π,所以A+C=2π3,故cos(A+C)=-12.答案C3.在等差数列{an}中,已知a1=13,a4+a5=163,ak=33,则k=()A.50B.49C.48D.47解析设等差数列{an}的公差为d, a1=13,a4+a5=163,∴2a1+7d=163,解得d=23,则an=13+(n-1)×23=2n-13,则ak=2k-13=33,解得k=50.答案A4.在等差数列{an}中,a1=8,a5=2,若在相邻两项之间各插入一个数,使之成等差数列,则新等差数列的公差为()A.34B.-34C.-67D.-1解析设原等差数列的公差为d,则8+4d=2,解得d=-32,因此新等差数列的公差为-34.答案B5.(多选)等差数列20,17,14,11,…中的负数项可以是()A.第7项B.第8项C.第9项D.第10项解析 a1=20,d=-3,∴an=20+(n-1)×(-3)=23-3n,∴a7=2>0,a8=-1<0.故数列中的负数项是第8项及其之后的项,故选BCD.答案BCD6.已知{an}为等差数列,若a2=2a3+1,a4=2a3+7,则a3=.解析 {an}为等差数列,a2=2a3+1,a4=2a3+7,∴a1+d=2(a1+2d)+1,a1+3d=2(a1+2d)+7,解得a1=-10,d=3,∴a3=a1+2d=-10+6=-4.答案-47.已知a>0,b>0,2a=3b=m,且a,ab,b成等差数列,则m=.解析 a>0,b>0,2a=3b=m≠1,∴a=lgmlg2,b=lgmlg3. a,ab,b成等差数列,∴2ab=a+b,∴2×lgmlg2×lgmlg3=lgmlg2+lgmlg3.∴lgm=12(lg2+lg3)=12lg6=lg√6.则m=√6.答案√68.正项数列{an}满足a1=1,a2=2,2an2=an+12+an-12(n∈N*,n≥2),则a7=.解析因为2an2=an+12+an-12(n∈N*,n≥2),所以数列{an2}是以a12=1为首项,以d=a22−a12=4-1=3为公差的等差数列,所以an2=1+3(n-1)=3n-2,所以an=√3n-2,n≥1.所以a7=√3×7-2=√19.答案√199.已知x,y,z成等差数列,求证:x2(y+z),y2(x+z),z2(y+x)也成等差数列.证明因为x,y,z成等差数列,所以2y=x+z,而x2(y+z)+z2(y+x)=x2y+x2z+z2y+z2x=x2y+z2y+xz(x+z)=x2y+z2y+2xyz=y(x+z)2=2y2(x+z),故x2(y+z),y2(x+z),z2(y+x)也成等差数列.10.已知数列{an},a1=1,an+1=2an+2n.(1)设bn=an2n-1,证明:数列{bn}是等差数列;(2)求数列{an}的通项公式.解(1)因为an+1=2an+2n,所以an+12n=2an+2n2n=an2n-1+1,所以an+12n−an2n-1=1,n∈N*.又因为bn=an2n-1,所以bn+1-bn=1.所以数列{bn}是等差数列,其首项b1=a1=1,公差为1.(2)由(1)知bn=1+(n-1)×1=n,所以an=2n-1bn=n·2n-1.能力提升练1.已知等差数列的前4项分别是a,x,b,2x,则ab等于()A.14B.12C.13D.23解析依题意,得{a+2x=x+b,2b=x+2x,解得{a=x2,b=3x2,故ab=13.答案C2.下列命题正确的是()A.若a,b,c成等差数列,则a2,b2,c2成等差数列B.若a,b,c成等差数列,则log2a,log2b,log2c成等差数列C.若a,b,c成等差数列,则a+2,b+2,c+2成等差数列D.若a,b,c成等差数列,则2a,2b,2c成等差数列解析因为a,b,c为等差数列,所以2b=a+c,所以2(b+2)=(a+2)+(c+2),故a+2,b+2,c+2成等差数列,即C项正确.ABD三项通过举反例易知不正确.答案C3.已知等差数列{an}满足4a3=3a2,则{an}中一定为零的项是()A.a6B.a8C.a10D.a12解析设等差数列{an}的公差为d. 4a3=3a2,∴4(a1+2d)=3(a1+d),可得a1+5d=0,∴a6=0,则{an}中一定为零的项是a6.答案A4.已知{an}是公差为d的等差数列,若3a6=a3+a4+a5+12,则d=.解析3a6=a3+a4+a5+12⇒3(a1+5d)=a1+2d+a1+3d+a1+4d+12⇒6d=12,解得d=2.答案25.已知数列{an}与{an2n}均为等差数列(n∈N*),且a1=1,则a10=.解析设等差数列{an}的公差为d,则an=1+(n-1)d=dn+1-d,∴an2n=d2n+2d(1-d)+(1-d)2n为等差数列,根据等差数列的性质可知(1-d)2n=0,即d=1,∴a10=10.答案106.已知数列{an},a1=1,a2=23,且1an-1+1an+1=2an(n≥2),则an=.解析 1an-1+1an+1=2an,∴数列{1an}是等差数列,公差d=1a2−1a1=12.∴1an=1a1+(n-1)d=1+12(n-1)=n+12.∴an=2n+1.答案2n+17.已知等差数列{an}:3,7,11,15,….(1)求等差数列{an}的通项公式.(2)135,4b+19(b∈N*)是数列{an}中的项吗?若是,是第几项?(3)若am,at(m,t∈N*)是数列{an}中的项,则2am+3at是数...
