第2课时等差数列的性质及应用课后篇巩固探究A组1.已知等差数列{an}中,a7+a9=16,a4=1,则a12的值是()A.15B.30C.31D.64解析: {an}是等差数列,∴a7+a9=a4+a12,∴a12=16-1=15.答案:A2.已知{an}为等差数列,a1+a3+a5=105,a2+a4+a6=99,则a20等于()A.-1B.1C.3D.7解析: a1+a3+a5=105,∴3a3=105,解得a3=35,同理由a2+a4+a6=99,得a4=33. d=a4-a3=33-35=-2,∴a20=a4+(20-4)d=33+16×(-2)=1.答案:B3.若{an}是等差数列,则下列数列中仍为等差数列的有()①{an+3}②{an2}③{an+1-an}④{2an}⑤{2an+n}A.1个B.2个C.3个D.4个解析:根据等差数列的定义判断,若{an}是等差数列,则{an+3},{an+1-an},{2an},{2an+n}均为等差数列,而{an2}不一定是等差数列.答案:D4.已知等差数列{an}满足a1+a2+a3+…+a101=0,则有()A.a1+a101>0B.a2+a100<0C.a3+a100≤0D.a51=0解析:由题设a1+a2+a3+…+a101=101a51=0,得a51=0.答案:D5.若等差数列的前三项依次是x-1,x+1,2x+3,则其通项公式为()A.an=2n-5B.an=2n-3C.an=2n-1D.an=2n+1解析: x-1,x+1,2x+3是等差数列的前三项,∴2(x+1)=x-1+2x+3,解得x=0.∴a1=x-1=-1,a2=1,a3=3,∴d=2.∴an=-1+2(n-1)=2n-3,故选B.答案:B16.在等差数列{an}中,a1+a4+a7=39,a2+a5+a8=33,则a3+a6+a9=.解析:由等差数列的性质,得(a1+a4+a7)+(a3+a6+a9)=2(a2+a5+a8),即39+(a3+a6+a9)=2×33,故a3+a6+a9=66-39=27.答案:277.若lg2,lg(2x-1),lg(2x+3)成等差数列,则x的值是.解析:由题意,知2lg(2x-1)=lg2+lg(2x+3),则(2x-1)2=2(2x+3),即(2x)2-4·2x-5=0,∴(2x-5)(2x+1)=0,∴2x=5,∴x=log25.答案:log258.已知一个等差数列由三个数构成,这三个数之和为9,平方和为35,则这三个数构成的等差数列为.答案:1,3,5或5,3,19.在等差数列{an}中,a1+a4+a7=15,a2a4a6=45,求数列{an}的通项公式.解 a1+a7=2a4=a2+a6,∴a1+a4+a7=3a4=15,∴a4=5,∴a2+a6=10,a2a6=9.∴a2,a6是方程x2-10x+9=0的两根.∴{a2=1,a6=9或{a2=9,a6=1.若a2=1,a6=9,则d=a6-a26-2=2,∴an=2n-3.若a2=9,a6=1,则d=a6-a26-2=-2,∴an=13-2n.∴数列{an}的通项公式为an=2n-3或an=13-2n.10.已知f(x)=x2-2x-3,等差数列{an}中,a1=f(x-1),a2=-32,a3=f(x),求:(1)x的值;(2)通项an.解(1)由f(x)=x2-2x-3,得a1=f(x-1)=(x-1)2-2(x-1)-3=x2-4x,a3=x2-2x-3,又因为{an}为等差数列,所以2a2=a1+a3,即-3=x2-4x+x2-2x-3,解得x=0或x=3.(2)当x=0时,a1=0,d=a2-a1=-32,2此时an=a1+(n-1)d=-32(n-1);当x=3时,a1=-3,d=a2-a1=32,此时an=a1+(n-1)d=32(n-3).B组1.在数列{an}中,若a2=2,a6=0,且数列{1an+1}是等差数列,则a4等于()A.12B.13C.14D.16解析:令bn=1an+1,则b2=1a2+1=13,b6=1a6+1=1.由题意知{bn}是等差数列,∴b6-b2=(6-2)d=4d=23,∴d=16.∴b4=b2+2d=13+2×16=23. b4=1a4+1,∴a4=12.答案:A2.已知数列{an}为等差数列,且a1+a7+a13=4π,则tan(a2+a12)的值为()A.√3B.±√3C.-√33D.-√3解析: {an}为等差数列,∴a1+a7+a13=3a7=4π.∴a7=4π3,tan(a2+a12)=tan2a7=tan8π3=-√3.答案:D3.《九章算术》“竹九节”问题:现有一根9节的竹子,自上而下各节的容积成等差数列,上面4节的容积共3升,下面3节的容积共4升,则第5节的容积为()A.1升B.6766升C.4744升D.3733升解析:设所构成的等差数列{an}的首项为a1,公差为d,由题意得{a1+a2+a3+a4=3,a7+a8+a9=4,即{4a1+6d=3,3a1+21d=4,3解得{a1=1322,d=766,所以a5=a1+4d=6766.答案:B4.导学号33194007在等差数列{an}中,如果a2+a5+a8=9,那么关于x的方程x2+(a4+a6)x+10=0()A.无实根B.有两个相等实根C.有两个不等实根D.不能确定有无实根解析: a4+a6=a2+a8=2a5,即3a5=9,∴a5=3.又a4+a6=2a5=6,∴关于x的方程为x2+6x+10=0,则判别式Δ=62-4×10<0,∴无实数解.答案:A5.已知logab,-1,logba成等差数列,且a,b为关于x的方程x2-cx+d=0的两根,则d=.解析:由已知,得logab+logba=-2,即lgblga+lgalgb=-2,从而有(lga+lgb)2=0,可得lga=-lgb=lg1b,即ab=1.故由根与系数的关系得d=ab=1.答案:16.导学号33194008已知方程(x2-2x+m)(x2-2x+n)=0的四个根组成一个首项为14的等差数列,则|m-n|=.解析:由题意设这4个根为14,14+d,14+2d,14+3d.可得14+(14+3d)=2,∴d=12.∴这4个根依次为14,34,54,74.∴n=14×74...
