1、考点规范练29等差数列及其前n项和考点规范练A册第18页基础巩固组1.若数列an的首项a1=1,且an=an-1+2(n2),则a7等于() A.13B.14C.15D.17答案:A解析:an=an-1+2(n2),an-an-1=2.又a1=1,数列an是以1为首项,以2为公差的等差数列,故a7=1+2(7-1)=13.2.已知Sn为等差数列an的前n项和,a2+a8=6,则S9等于()A.272B.27C.54D.108答案:B解析:S9=9(a1+a9)2=9(a2+a8)2=27.3.(2015北京,理6)设an是等差数列.下列结论中正确的是()A.若a1+a20,则a2+a30B.若
2、a1+a30,则a1+a20C.若0a1a1a3D.若a10答案:C解析:设等差数列公差为d.对于A选项,a1+a2=2a1+d0,而a2+a3=2a1+3d不一定大于0;对于B选项,a1+a3=2a1+2d0,a1+a2=2a1+d不一定小于0;对于C选项,0a10,故a2=a1+a32a1a3;对于D选项,(a2-a1)(a2-a3)=-d20.故只有C正确.4.在等差数列an中,a2=3,a3+a4=9,则a1a6的值为()A.14B.18C.21D.27答案:A解析:设等差数列an的公差为d,则依题意得a1+d=3,2a1+5d=9,由此解得a1=2,d=1,故a6=a1+5d=7,即
3、a1a6=14.5.已知每项均大于零的数列an中,首项a1=1,且前n项和Sn满足SnSn-1-Sn-1Sn=2SnSn-1(nN+,且n2),则a81等于()A.638B.639C.640D.641导学号92950488答案:C解析:由已知SnSn-1-Sn-1Sn=2SnSn-1,可得Sn-Sn-1=2,Sn是以1为首项,2为公差的等差数列,故Sn=2n-1,Sn=(2n-1)2,a81=S81-S80=1612-1592=640,故选C.6.(2015广州综合测试)设Sn是等差数列an的前n项和,公差d0,若S11=132,a3+ak=24,则正整数k的值为()A.9B.10C.11D.
4、12答案:A解析:依题意得S11=11(a1+a11)2=11a6=132,a6=12,于是有a3+ak=24=2a6,因此3+k=26=12,k=9,故选A.7.已知数列an是等差数列,a1+a3+a5=105,a2+a4+a6=99,an的前n项和为Sn,则使得Sn达到最大的n是()A.18B.19C.20D.21答案:C解析:a1+a3+a5=105a3=35,a2+a4+a6=99a4=33,则an的公差d=33-35=-2,a1=a3-2d=39,Sn=-n2+40n,因此当Sn取得最大值时,n=20.8.已知等差数列an的前n项和为Sn,且S10=10,S20=30,则S30=.答
5、案:60解析:Sn是等差数列an的前n项和,S10,S20-S10,S30-S20也成等差数列.2(S20-S10)=S10+(S30-S20).S30=60.9.(2015江苏无锡一模)已知数列an中,a1=1,a2=2,当整数n2时,Sn+1+Sn-1=2(Sn+S1)都成立,则S15=.答案:211解析:由Sn+1+Sn-1=2(Sn+S1)得(Sn+1-Sn)-(Sn-Sn-1)=2S1=2,即an+1-an=2(n2),数列an从第二项起构成以2为首项,2为公差的等差数列,则S15=1+214+141322=211.10.在等差数列an中,a1=7,公差为d,前n项和为Sn,当且仅当
6、n=8时,Sn取得最大值,则d的取值范围为.答案:-1,-78解析:由题意知当d0,数列an中所有非负项的和最大.又当且仅当n=8时,Sn取最大值,a80,a90,7+8d0,解得-1d0,a3203时,f(n)0,0n203时,f(n)0,当n=203时,f(n)取最小值,而nN+,则f(6)=-48,f(7)=-49,当n=7时,f(n)取最小值-49.17.(2015安徽宿州调研)已知函数f(x)=x2-2(n+1)x+n2+5n-7.(1)设函数y=f(x)的图像的顶点的纵坐标构成数列an,求证:an为等差数列;(2)设函数y=f(x)的图像的顶点到x轴的距离构成数列bn,求bn的前n项和Sn.解:(1)证明:f(x)=x2-2(n+1)x+n2+5n-7=x-(n+1)2+3n-8,an=3n-8,an+1-an=3(n+1)-8-(3n-8)=3,数列an为等差数列.(2)由题意知,bn=|an|=|3n-8|,当1n2时,bn=8-3n,Sn=b1+bn=n(b1+bn)2=n5+(8-3n)2=13n-3n22;当n3时,bn=3n-8,Sn=b1+b2+b3+bn=5+2+1+(3n-8)=7+(n-2)1+(3n-8)2=3n2-13n+282.Sn=13n-3n22,1n2,3n2-12n+282,n3.导学号929504934
