新版高中數(shù)學(xué)人教A版必修5習(xí)題第二章數(shù)列2.2.2

第2課時(shí)等差數(shù)列的性質(zhì)課時(shí)過關(guān)·能力提升基礎(chǔ)鞏固1已知m和2n的等差中項(xiàng)是4,2m和n的等差中項(xiàng)是5,則m和n的等差中項(xiàng)是().A.2 B.3 C.6 D.9解析:∵由已知可得m+2n=8,2m+n=10,∴3(m+n)=18,∴m+n=6.∴m和n的等差中項(xiàng)是3.故選B.答案:B2若等差數(shù)列{an}的公差為d,則數(shù)列{can}(c為常數(shù),且c≠0)是().A.公差為d的等差數(shù)列 B.公差為cd的等差數(shù)列C.不是等差數(shù)列 D.以上都不對(duì)解析:設(shè)bn=can,則bn+1bn=can+1can=c(an+1an)=cd.答案:B3已知等差數(shù)列{an}的公差為d(d≠0),且a3+a6+a10+a13=32,若am=8,則m等于().A.12 B.8 C.6 D.4解析:∵a3+a6+a10+a13=(a3+a13)+(a6+a10)=2a8+2a8=4a8=32,∴a8=8.∴m=8.答案:B4若數(shù)列{an}是等差數(shù)列,a15=8,a60=20,則a75的值為().A.12 B.16 C.24 D.48解析:∵{an}是等差數(shù)列,∴a15,a30,a45,a60,a75成等差數(shù)列.設(shè)其公差為D,則a60=a15+3D,即D=4,故a75=a15+4D=8+4×4=24.答案:C5已知數(shù)列{an}是等差數(shù)列,若a1a5+a9a13+a17=117,則a3+a15=.

解析:a1a5+a9a13+a17=(a1+a17)(a5+a13)+a9=a9=117,a3+a15=2a9=2×117=234.答案:2346在數(shù)列{an}中,a1,a12是方程x2-2解析:由題意得a1+a12=故a5+a8=a1+a12=答案:27已知在數(shù)列{an}中,a5=10,a12=31,則其公差d=.

解析:d=答案:38在等差數(shù)列{an}中,已知a5=10,a12>31,求公差d的取值范圍.解設(shè)首項(xiàng)為a1,由題意,可知解得d>3.所以d的取值范圍是(3,+∞).9已知三個(gè)數(shù)成等差數(shù)列,其和為15,首末兩項(xiàng)的積為9,求這三個(gè)數(shù).解由題意,可設(shè)這三個(gè)數(shù)分別為ad,a,a+d,則解得所以,當(dāng)d=4時(shí),這三個(gè)數(shù)為1,5,9;當(dāng)d=4時(shí),這三個(gè)數(shù)為9,5,1.所以這三個(gè)數(shù)為1,5,9或9,5,1.10已知數(shù)列{an},an=2n1,bn=a2n1.(1)求{bn}的通項(xiàng)公式;(2)數(shù)列{bn}是否為等差數(shù)列?說明理由.解(1)∵an=2n1,bn=a2n1,∴bn=a2n1=2(2n1)1=4n3.(2){bn}是等差數(shù)列.理由如下:由bn=4n3,知當(dāng)n≥2時(shí),bn1=4(n1)3=4n7.∴bnbn1=(4n3)(4n7)=4.又b1=a1=2×11=1,∴{bn}是首項(xiàng)b1=1,公差為4的等差數(shù)列.能力提升1設(shè)數(shù)列{an},{bn}都是等差數(shù)列,且a1=25,b1=75,a2+b2=100,則a37+b37等于().A.0 B.37 C.100 D.37解析:∵{an},{bn}都是等差數(shù)列,∴數(shù)列{an+bn}也是等差數(shù)列,設(shè)其公差為d,則d=(a2+b2)(a1+b1)=0.∴數(shù)列{an+bn}為常數(shù)列.∴a37+b37=a1+b1=100.答案:C2在如圖所示的表格里,每格填上一個(gè)數(shù)字后使每一橫行和豎列都成等差數(shù)列,則a等于().A.3 B.3 C.0 D.6解析:由于第一行成等差數(shù)列,則第一行中間數(shù)為2+62=4.又第二列成等差數(shù)列,則4+a=2×答案:C3如果在等差數(shù)列{an}中,a5+a6+a7=15,那么a3+a4+…+a9等于().A.21 B.30 C.35 D.40解析:a5+a6+a7=(a5+a7)+a6=2a6+a6=3a6=15,所以a6=5.所以a3+a4+…+a9=(a3+a9)+(a4+a8)+(a5+a7)+a6=7a6=35.答案:C4在等差數(shù)列{an}中,a3+a7a10=8,a11a4=4,則a6+a7+a8等于().A.34 B.35 C.36 D.37解析:由題意得(a3+a7a10)+(a11a4)=12,∴(a3+a11)+a7(a10+a4)=12.∵a3+a11=a10+a4,∴a7=12.∴a6+a7+a8=3a7=36.答案:C5若等差數(shù)列{an}滿足a7+a8+a9>0,a7+a10<0,則當(dāng)n=時(shí),{an}的前n項(xiàng)和最大.

答案:86若a,b,c成等差數(shù)列,則二次函數(shù)y=ax22bx+c的圖象與x軸的交點(diǎn)的個(gè)數(shù)為.

解析:∵a,b,c成等差數(shù)列,∴2b=a+c.∴Δ=4b24ac=(a+c)24ac=(ac)2≥0.∴二次函數(shù)y=ax22bx+c的圖象與x軸的交點(diǎn)個(gè)數(shù)為1或2.答案:1或2★7已知在數(shù)列{an}中,a3=3,a7=1,又?jǐn)?shù)列1解析:∵1an+1是等差數(shù)列則b3=∴公差d=∴bn=b3+∴∴an+1=答案:158已知1證明∵1a∴2ac=ab+bc.∴2ac=2ac2b(a+c),∴2ac+a2+c2=2ac2b(a+c)+a2+c2,∴(ac)2=(a+c)(a+c2b).∵ac,a+c,a+c2b都是正數(shù),∴2lg(ac)=lg(a+c)+lg(a+c2b).∴l(xiāng)g(a+c),lg(ac),lg(a+c2b)成等差數(shù)列.★9設(shè)數(shù)列{an}是等差數(shù)列,bn=1解∵b1b2b3=18,∴∴∴a1+a2+a3=3.又{an}成等差數(shù)列,∴a