（新教材適用）高中數(shù)學(xué)第四章數(shù)列42等差數(shù)列421等差數(shù)列的概念第1課時(shí)等差數(shù)列的通項(xiàng)公式課后習(xí)題新人教A版選擇性

4.2.1等差數(shù)列的概念第1課時(shí)等差數(shù)列的通項(xiàng)公式課后訓(xùn)練鞏固提升A組1.等差數(shù)列1,1,3,…,89共有()項(xiàng).A.92 B.47 C.46 D.45解析:由題意知首項(xiàng)a1=1,d=2,故89=1+(n1)×(2),解得n=46.答案:C2.(多選題)下列說(shuō)法中正確的有()A.若a,b,c成等差數(shù)列,則a2,b2,c2一定成等差數(shù)列B.若a,b,c成等差數(shù)列,則2a,2b,2c可能成等差數(shù)列C.若a,b,c成等差數(shù)列,則ka+2,kb+2,kc+2(k為常數(shù))一定成等差數(shù)列D.若a,b,c成等差數(shù)列,則1a解析:對(duì)于A,取a=1,b=2,c=3,可得a2=1,b2=4,c2=9,顯然,a2,b2,c2不成等差數(shù)列,故A錯(cuò)誤;對(duì)于B,取a=b=c,可得2a=2b=2c,故B正確;對(duì)于C,因?yàn)閍,b,c成等差數(shù)列,所以a+c=2b.所以(ka+2)+(kc+2)=k(a+c)+4=2(kb+2),即ka+2,kb+2,kc+2(k為常數(shù))成等差數(shù)列,故C正確;對(duì)于D,a=b=c≠0?1a=綜上可知,B,C,D正確.答案:BCD3.已知數(shù)列{an}為等差數(shù)列,且a1=2,a3=8,則a4+a6=()A.31 B.29 C.28 D.26解析:依題意,數(shù)列{an}為等差數(shù)列,且a1=2,a3=8,設(shè)d為公差,所以2d=a3a1=82=6,所以d=3,所以a4+a6=2a1+8d=4+24=28.答案:C4.已知等差數(shù)列{an}中,a7+a9=16,a4=1,則a12的值是()A.15 B.30 C.31 D.64解析:設(shè)數(shù)列{an}的首項(xiàng)為a1,公差為d,則有a1+6故a12=a1+11d=15.答案:A5.若x是a,b的等差中項(xiàng),x2是a2,b2的等差中項(xiàng),則a與b的關(guān)系為()A.a=b=0 B.a=bC.a=3b D.a=b或a=3b解析:由題意得2x=a+b,故3b2+2aba2=0,即a=b或a=3b.答案:D6.已知等差數(shù)列{an}滿足a1=0,a5=4,則公差d=,a2+a4=.

解析:因?yàn)榈炔顢?shù)列{an}滿足a1=0,a5=4,所以4=0+4d,則公差d=1,a2+a4=2a1+4d=4.答案:147.已知等差數(shù)列{an}中,a1+a3+a9=20,則4a5a7=.

解析:等差數(shù)列{an}中,a1+a3+a9=3a1+10d=20,則4a5a7=4(a1+4d)(a1+6d)=3a1+10d=20.答案:208.在等差數(shù)列{an}中,若ap=q,aq=p,則ap+q=.

解析:設(shè)數(shù)列{an}的公差為d,∵ap=a1+(p1)d,aq=a1+(q1)d,∴d=q-pp-q=1,∴ap+q=a1+(p+q1)d=0.答案:09.在公差為d的等差數(shù)列{an}中,(1)已知a1=5,d=2,求a10;(2)已知a1=3,d=4,an=59,求n;(3)已知d=34,a28=1414,求a(4)已知a5=21,a10=36,求a1和d.解:(1)a10=a1+(101)d=5+9×2=23.(2)∵an=a1+(n1)d,∴59=3+4(n1),解得n=15.(3)∵a28=a1+27d,∴a1=a2827d=141427×34=(4)由a5=10.已知數(shù)列{an}滿足a1=2,an+1=2a(1)數(shù)列1an(2)求an.解:(1)數(shù)列1an是等差數(shù)列∵a1=2,an+1=2a∴1an+1=即1an是首項(xiàng)為1a1=(2)由(1)可知1an=1a1+(n1)d=nB組1.在等差數(shù)列{an}中,若a1011=5,a1+2a4=9,則a2019=()A.9 B.8 C.7 D.6解析:設(shè)數(shù)列{an}的公差為d,a1011=a1+1010d=5,a1+2a4=3a1+6d=9,解得1008d=2,故a2019=a1+2018d=a1+1010d+1008d=5+2=7.答案:C2.設(shè)等差數(shù)列{an}的公差為d.若數(shù)列{2a1an}為遞減數(shù)列,則()A.d>0 B.d<0 C.a1d>0 D.a1d<0解析:設(shè)bn=2a1an,則bn+1=2a1an+1,由于{2a1an}是遞減數(shù)列,則bn>bn+1,即2a1an>2a1an+1.∴a1an>a1an+1,∴a1ana1(an+d)>0,∴a1(anand)>0,即a1(d)>0,∴a1d<0.答案:D3.已知等差數(shù)列{an}的首項(xiàng)為70,公差為9,則這個(gè)數(shù)列中絕對(duì)值最小的一項(xiàng)為()A.a8 B.a9 C.a10 D.a11解析:∵an=70+(n1)×(9)=799n,∴a8=7,a9=2,a10=11,故絕對(duì)值最小的一項(xiàng)為a9.答案:B4.已知首項(xiàng)為24的等差數(shù)列,從第10項(xiàng)開(kāi)始為正數(shù),則公差d的取值范圍是()A.83,+∞ BC.83,3解析:由題意得a10=-24+9d答案:C5.(多選題)已知等差數(shù)列{an}的公差d>0,則下列說(shuō)法正確的是()A.數(shù)列{an}是遞增數(shù)列B.數(shù)列{nan}是遞增數(shù)列C.數(shù)列anD.數(shù)列{an+3nd}是遞增數(shù)列解析:因?yàn)閷?duì)于公差d>0的等差數(shù)列{an},an+1an=d>0,所以數(shù)列{an}是遞增數(shù)列成立,A正確;對(duì)于數(shù)列{nan},第n+1項(xiàng)與第n項(xiàng)的差為(n+1)·an+1nan=(n+1)d+an,不一定是正實(shí)數(shù),B錯(cuò)誤;對(duì)于數(shù)列ann,第n+1項(xiàng)與第n項(xiàng)的差為對(duì)于數(shù)列{an+3nd},第n+1項(xiàng)與第n項(xiàng)的差為an+1+3(n+1)dan3nd=4d>0,故數(shù)列{an+3nd}是遞增數(shù)列成立,D正確.故選AD.答案:AD6.已知正項(xiàng)數(shù)列{an}滿足a1=1,a2=2,2an2=an+12+a解析:因?yàn)?an2=所以數(shù)列{an2}是以a12=1為首項(xiàng),以d=a所以an2=1+3(n1)=3所以an=3n-2,所以a7=3×答案:197.設(shè)a>0,b>0,若lga與lgb的等差中項(xiàng)為0,則1a+1b解析:∵a>0,b>0,lga與lgb的等差中項(xiàng)為0,∴l(xiāng)ga+lgb=2×0,解得ab=1.則1a+1b≥21ab∴1a+答案:28.已知1b+c,1c+a,1a證明:∵1b∴2c∴2(b+c)(a+b)=(a+c)(a+c+2b),∴2b2=a2+c2,∴a2,b2,c2成等差數(shù)列.9.已知數(shù)列{an}滿足a1=4,an=44an-1(n≥2