1、等比数列复习课课时过关能力提升1.在等比数列an中,a2=9,a5=243,则an的前4项和为()A.81B.120C.168D.192解析:q3=a5a2=2439=27,q=3.a1=3.S4=3(1-34)1-3=120.答案:B2.Sn为等差数列an的前n项和,S9=-36,S13=-104,等比数列bn中,b5=a5,b7=a7,则b6等于()A.42B.-42C.42D.22解析:由S9=9a5=-36,得a5=-4,b5=-4.由S13=13a7=-104,得a7=-8,b7=-8.b62=b5b7=32.b6=42.答案:C3.已知各项不为0的等差数列an满足a4-2a72+3
2、a8=0,数列bn是等比数列,且b7=a7,则b2b8b11等于()A.1B.2C.4D.8解析:设等差数列的公差为d,由a4-2a72+3a8=0,得a7-3d-2a72+3(a7+d)=0,解得a7=2或a7=0(因为a7=b7,而bn是等比数列,所以a7=0舍去).设数列bn的公比为q,则b7=a7=2,所以b2b8b11=b52b11=b72b11q4=b73=23=8.答案:D4.一个等比数列共有3m项,若前2m项之和为15,后2m项之和为60,则这个等比数列的所有项的和为()A.63B.72C.75D.87解析:由已知S2m=15,S3m-Sm=60,又(S2m-Sm)2=Sm(S
3、3m-S2m)=Sm(Sm+60-S2m),解得Sm=3,所以S3m=60+3=63.答案:A5.已知a,b,c为等比数列,b,m,a和b,n,c是两个等差数列,则am+cn等于()A.4B.3C.2D.1解析:因为b,m,a和b,n,c是两个等差数列,所以m=a+b2,n=b+c2.又a,b,c为等比数列,所以b2=ac.所以am+cn=2aa+b+2cb+c=2ab+2ac+2ac+2bc(a+b)(b+c)=2ab+4ac+2bcab+ac+b2+bc=2(ab+2ac+bc)ab+2ac+bc=2.答案:C6.数列an是各项为正数的等比数列,且a4=2,已知函数f(x)=log12x,
4、则f(a13)+f(a23)+f(a73)=()A.-6B.-21C.-12D.21解析:f(a13)+f(a23)+f(a73)=log12(a13a23a73)=log12(a1a2a7)3=log12(a47)3=21log12a4=21log122=-21.答案:B7.已知数列an是等差数列,若a2,a4+3,a6+6构成公比为q的等比数列,则q等于()A.1B.2C.3D.4解析:令等差数列an的公差为d,由a2,a4+3,a6+6构成公比为q的等比数列,得(a4+3)2=a2(a6+6),即(a1+3d+3)2=(a1+d)(a1+5d+6),所以(2d+3)2=0,即d=-32.
5、所以q=a4+3a2=a1-92+3a1-32=a1-32a1-32=1,故选A.答案:A8.设等比数列an的公比为q,前n项和为Sn,若Sn+1,Sn,Sn+2成等差数列,则q为.答案:-29.已知等比数列an,前n项和为Sn,a1+a2=34,a4+a5=6,则S6=.解析:记等比数列an的公比为q,则有q3=a4+a5a1+a2=8,q=2,S6=(a1+a2)+q2(a1+a2)+q4(a1+a2)=21(a1+a2)=634.答案:63410.数列an的前n项和为Sn=n2+n+1,bn=(-1)nan(nN+),则数列bn的前50项和为.解析:Sn=n2+n+1,当n=1时,a1=
6、S1=3;当n2时,an=Sn-Sn-1=2n,an=3,n=1,2n,n2.bn=(-1)nan,T50=-3+4-6+8-98+100=49.答案:4911.若a,b,c成等差数列,且a+1,b,c与a,b,c+2都成等比数列,求b的值.解:设a,b,c分别为b-d,b,b+d.由题意得b-d+1,b,b+d与b-d,b,b+d+2都成等比数列,所以b2=(b-d+1)(b+d),b2=(b-d)(b+d+2),整理,得b2=b2-d2+b+d,b2=b2-d2+2b-2d,所以b+d=2b-2d,即b=3d,代入,得9d2=(3d-d+1)(3d+d),解得d=4或d=0.当d=0时,a
7、=b=c,此时a+1,b,c不成等比数列.所以d0,d=4.所以b=12.12.已知公差不为0的等差数列an的首项a1=a,a0,前n项和为Sn,且1a1,1a2,1a4成等比数列.(1)求数列an的通项公式;(2)设数列1Sn的前n项和为An,若A2 017=2 0172 018,求a的值.解:(1)设等差数列an的公差为d,由1a22=1a11a4,即a22=a1a4,得(a1+d)2=a1(a1+3d).因为d0,所以d=a1=a.所以an=a+(n-1)a=na.(2)因为Sn=n(a1+an)2=n(n+1)a2,所以1Sn=2a1n-1n+1.所以An=1S1+1S2+1S3+1Sn=2a11-12+12-13+13-14+1n-1n+1=2a1-1n+1.又A2 017=2a1-12 018=2 0172 018,所以a=2.
