人教版導(dǎo)與練總復(fù)習(xí)數(shù)學(xué)一輪課時(shí)作業(yè)：第五章第2節(jié)　等差數(shù)列及其前n項(xiàng)和

1、第2節(jié)等差數(shù)列及其前n項(xiàng)和 知識(shí)點(diǎn)、方法基礎(chǔ)鞏固練綜合運(yùn)用練應(yīng)用創(chuàng)新練等差數(shù)列的基本量運(yùn)算1,6,7等差數(shù)列的判定與證明2,410,15,17等差數(shù)列的性質(zhì)3,8等差數(shù)列前n項(xiàng)和的最值512,1418等差數(shù)列的綜合應(yīng)用911,13,161.已知等差數(shù)列an中,a2=1,前5項(xiàng)和S5=-15,則數(shù)列an的公差為(D)A.-3B.-52C.-2D.-42.若等差數(shù)列an的公差為d,則數(shù)列a2n-1是(B)A.公差為d的等差數(shù)列B.公差為2d的等差數(shù)列C.公差為nd的等差數(shù)列D.非等差數(shù)列解析:數(shù)列a2n-1其實(shí)就是a1,a3,a5,a7,奇數(shù)項(xiàng)組成的數(shù)列,它們之間相差2d.故選B.3.設(shè)Sn是等差

2、數(shù)列an的前n項(xiàng)和,若a6a5=911,則S11S9等于(A)A.1B.-1C.2D.12解析:S11S9=(a1+a11)211(a1+a9)29=11a69a5=1.故選A.4.已知Sn是數(shù)列an的前n項(xiàng)和,且Sn+1=Sn+an+3,a4+a5=23,則S8等于(C)A.72B.88C.92D.98解析:由Sn+1=Sn+an+3,得an+1-an=3,故數(shù)列an是公差為3的等差數(shù)列.又a4+a5=23,所以2a1+7d=2a1+21=23,所以a1=1,所以S8=8a1+872d=92.故選C.5.在等差數(shù)列an中,已知a3+a80,且S90,則S1,S2,S9中最小的是(A)A.S5

3、B.S6C.S7D.S8解析:在等差數(shù)列an中,S9=9(a1+a9)2=9a50,所以a50,所以a60,所以數(shù)列從第6項(xiàng)開始大于零,前5項(xiàng)都小于零,所以S5最小.故選A.6.已知數(shù)列an滿足an-an-1=k(n2,nN*),a1=k,則數(shù)列an的通項(xiàng)公式為.解析:因?yàn)閿?shù)列an滿足an-an-1=k(n2,nN*),所以數(shù)列an為公差為k的等差數(shù)列,又a1=k,所以an=a1+(n-1)k=k+(n-1)k,即an=nk.答案:an=nk7.設(shè)Sn為等差數(shù)列an的前n項(xiàng)和,滿足S2=S6,S55-S44=2,則a1=,公差d=.解析:由an為等差數(shù)列,得數(shù)列Snn是公差為d2的等差數(shù)列.因

4、為S55-S44=2,所以d2=2d=4.又S2=S62a1+4=6a1+6524a1=-14.答案:-1448.在等差數(shù)列an中,公差d=12,前100項(xiàng)的和S100=45,則a1+a3+a5+a99=.解析:因?yàn)镾100=1002(a1+a100)=45,所以a1+a100=910,則a1+a99=a1+a100-d=25,則a1+a3+a5+a99=502(a1+a99)=50225=10.答案:109.已知數(shù)列an是等差數(shù)列.(1)前四項(xiàng)和為21,末四項(xiàng)和為67,且各項(xiàng)和為286,求項(xiàng)數(shù);(2)項(xiàng)數(shù)為奇數(shù),奇數(shù)項(xiàng)和為44,偶數(shù)項(xiàng)和為33.求數(shù)列的中間項(xiàng)和項(xiàng)數(shù).解:(1)由已知得a1+a

5、2+a3+a4=21,an-3+an-2+an-1+an=67,所以a1+a2+a3+a4+an-3+an-2+an-1+an=88,所以a1+an=884=22.因?yàn)镾n=286,所以n(a1+an)2=286,所以11n=286,所以n=26.(2)法一設(shè)項(xiàng)數(shù)為2k+1,則a1+a3+a2k+1=44=k+12(a1+a2k+1),a2+a4+a2k=33=k2(a2+a2k),又因?yàn)閍1+a2k+1=a2+a2k,所以k+1k=4433,所以k=3,項(xiàng)數(shù)為7,所以中間項(xiàng)為a1+a2k+12=11.法二記等差數(shù)列an的中間項(xiàng)為a中,奇數(shù)項(xiàng)和為S奇,偶數(shù)項(xiàng)和為S偶,前n項(xiàng)和為Sn.根據(jù)題意得

6、S偶+S奇=Sn,S奇-S偶=a中,所以Sn=77,a中=11,又na中=Sn,所以n=7.10.已知無窮數(shù)列an是各項(xiàng)均為正數(shù),且公差不為零的等差數(shù)列,其前n項(xiàng)和為Sn,nN*,則(D)A.數(shù)列Snn不可能是等差數(shù)列B.數(shù)列Snn2不可能是等差數(shù)列C.數(shù)列Snan不可能是等差數(shù)列D.數(shù)列anSn不可能是等差數(shù)列解析:由題可知Sn=na1+n(n-1)d2,an=a1+(n-1)d,其中a10,d 0,對(duì)A,Snn=a1+n-12d,所以數(shù)列Snn是公差為d2的等差數(shù)列,故A錯(cuò)誤;對(duì)B,Snn2=a1n+n-12nd=d2+(a1-d2)n,當(dāng)a1=d2時(shí),Snn2=d2,所以數(shù)列Snn2可能

7、是等差數(shù)列,故B錯(cuò)誤;對(duì)C,Snan=na1+n(n-1)d2a1+(n-1)d,當(dāng)a1=d時(shí),Snan=n+12,所以數(shù)列Snan可能是等差數(shù)列,故C錯(cuò)誤;anSn=a1+(n-1)dna1+n(n-1)d2,anSn不可能轉(zhuǎn)化為關(guān)于n的一次函數(shù)形式,故數(shù)列anSn不可能是等差數(shù)列,故D正確.故選D.11.(多選題)已知數(shù)列an是公差不為0的等差數(shù)列,前n項(xiàng)和為Sn,滿足a1+5a3=S8,則下列選項(xiàng)正確的有(AC)A.a10=0B.S10最小C.S7=S12D.S20=0解析:根據(jù)題意,數(shù)列an是等差數(shù)列,若a1+5a3=S8,即a1+5a1+10d=8a1+28d,變形可得a1=-9d.

8、又由an=a1+(n-1)d=(n-10)d,則有a10=0,故A一定正確;不能確定a1和d的符號(hào),不能確定S10最小,故B不正確;又由Sn=na1+n(n-1)d2=-9nd+n(n-1)d2=d2(n2-19n),則有S7=S12,故C一定正確;則S20=20a1+20192d=-180d+190d=10d.因?yàn)閐0,所以S200,則D不正確.故選AC.12.(2021寧夏吳忠高三一模)數(shù)列an是等差數(shù)列,Sn為其前n項(xiàng)和,且a10,a2 020+a2 0210,a2 020a2 0210,則使Sn0成立的最大正整數(shù)n是(C)A.2 020B.2 021C.4 040D.4 041解析:設(shè)

9、數(shù)列an的公差為d,由a10,a2 020+a2 0210,a2 020a2 0210,可知a2 0200,所以d0,數(shù)列an為遞增數(shù)列,S4 041=4 041(a1+a4 041)2=4 041a2 0210,S4 040=2 020(a1+a4 040)=2 020(a2 020+a2 021)0,所以可知使Sn0成立的n的最大值為4 040.故選C.13.等差數(shù)列an,bn滿足:對(duì)任意的nN*,都有anbn=2n+34n-9,則a7b3+b9+a5b4+b8=.解析:由等差數(shù)列的性質(zhì)可得b3+b9=b4+b8=2b6,a7+a5=2a6.所以a7b3+b9+a5b4+b8=a7+a52

10、b6=2a62b6=a6b6=26+346-9=1.答案:114.在等差數(shù)列an中,a1=7,公差為d,前n項(xiàng)和為Sn,當(dāng)且僅當(dāng)n=8時(shí),Sn取得最大值,則d的取值范圍是.解析:因?yàn)楫?dāng)且僅當(dāng)n=8時(shí),Sn有最大值,所以d0,a90,即d0,7+8d0,解得-1d0,求使得Snan的n的取值范圍.解:(1)設(shè)an的公差為d.由S9=-a5得a1+4d=0.由a3=4,得a1+2d=4.于是a1=8,d=-2.因此an的通項(xiàng)公式為an=10-2n.(2)由(1)得a1=-4d,故an=(n-5)d,Sn=n(n-9)d2.由a10知d0,故Snan等價(jià)于n2-11n+100,解得1n10.所以n的取值范圍是n|1n10,nN*.17.已知數(shù)列an的首項(xiàng)a1=1,2anan+1=an-an+1(nN*).(1)證明:數(shù)列1an是等差數(shù)列;(2)設(shè)bn=anan+1,數(shù)列bn的前n項(xiàng)和為Sn,求證:Sn12.證明:(1)由于a1=1,2anan+1=an-an+1,顯然anan+10,所以兩邊同除以anan+1可得,1an+1-1an=2,所以數(shù)列1an是以