1、A組專項(xiàng)基礎(chǔ)訓(xùn)練(時(shí)間：35分鐘，滿分：57分一、選擇題(每小題5分，共20分1. 設(shè)等差數(shù)列an前n項(xiàng)和為Sn,若al = - 11, a4+ a6= 6，則當(dāng)Sn取最小值時(shí)，n等于(A . 6 B . 7 C. 8 D . 9答案 A解析設(shè)該數(shù)列的公差為d,則 a4 + a6= 2a1 + 8d= 2X( 11 +8d= 6,解得 d = 2,Sn = 11n+ X2= n2 12n= (n 62 36,當(dāng)n= 6時(shí)，取最小值.2. 已知an為等比數(shù)列，Sn是它的前n項(xiàng)和.若a2 a3= 2a1,且a4與2a7的等差中項(xiàng)為，貝US5等于(A . 35 B. 33 C. 31 D . 29答

2、案 C解析 設(shè)數(shù)列an的公比為q，則由等比數(shù)列的性質(zhì)知，a2 a3 = a 1 a4 = 2a1，即 a4= 2.由a4與2a7的等差中項(xiàng)為知，a4 + 2a7= 2 X,a 7 =q3= =，即卩 q =,.a4 = a1q3= al x= 2,- al = 16,. S5= 31.3. 數(shù)列an的前n項(xiàng)和為Sn,若al = 1, an+ 1= 3Sn(n貝U a6等于(A . 3X44 B . 3 >44 + 1C. 43 D. 43+ 1答案 A解析 由 an + 1 = 3Sn?Sn+1 Sn= 3Sn? Sn+ 1=4Sn,數(shù)列Sn是首項(xiàng)為1,公比為4的等比數(shù)列，Sn = 4n

3、 1 , a6= S6 S5 = 45 44= 3 X44.4. 已知等差數(shù)列an的公差d = 2, a1 + a4 + a7+ a97= 50,那么a3 + a6 + a9+ a99的值是(A . 78 B . 82 C. 148 D . 182答案 B解析 / a3+ a6 + a9+ a99= (a1 + 2d + (a4 +2d + (a7 + 2d + (a97 + 2d=a1 + a4 + a7 +a97 + 2dX33=50 + 66X( 2=82.二、填空題(每小題5分，共15分答案 10解析 設(shè)等差數(shù)列an的前n項(xiàng)和為Sn,則S9S4= 0,即 a5 + a6 + a7 +

4、a8+ a9 = 0,5a7 = 0,故 a7 = 0.而 ak+ a4= 0,故 k = 10.6 . 已知數(shù)列 an的前n項(xiàng)和為Sn ,且Sn = 2n an ,則數(shù)列an的通項(xiàng)公式an =(2設(shè)bn= 2an +2n,求數(shù)列bn的前n項(xiàng)和Tn.解(1設(shè)等差數(shù)列an的公差為d,由題意，得解得所以 an = 2n 1.(2 因?yàn)?bn = 2an + 2n=X4n+2n,所以 Tn = bl + b2 + + bn=(4 + 42+ 4n+ 2(1 + 2+ + n=+ n2+ n= X4n+ n2+ n.9. (12分已知數(shù)列an的前n項(xiàng)和為Sn,且滿足：an+2SnSr 1= 0(n&g

5、t;2 n N , al =，判斷 與a n是否為等差數(shù)列，并說(shuō)明你的理由.解 因?yàn)?an= Sn Sn 1(n >2又因?yàn)?an+ 2SnSn 1 = 0,所以 Sn Sn 1 + 2SnSn 1 = 0(n > 2所以一=2(n又因?yàn)镾1 = a1=,所以是以2為首項(xiàng)，2為公差的等差數(shù)列.所以=2+ (n 1 ><2 = 2n,故 Sn =.所以當(dāng) n>L 時(shí)，an= Sn Sn 1 =,所以an + 1=,而 an+ 1 an =列.綜上，可知是等差數(shù)列，an不是等差數(shù)列.B組專項(xiàng)能力提升(時(shí)間：25分鐘，滿分：43分一、選擇題(每小題5分，共15分1. 已

6、知數(shù)列an是首項(xiàng)為a 1 = 4的等比數(shù)列，且 4a 1, a5, 2a3成等差數(shù)列，則其公比 q等 于(A . 1 B . 1 C. 1 或1 D.答案C解析 依題意，有 2a5= 4a1 2a3,即 2a1q4 = 4a1 2a1q2,整理得q4+ q2 2 = 0，解得q2= 1(q2 = 2舍去，所以q = 1或q= 1.2. 已知函數(shù)f(x =把函數(shù)g(x= f(x x的零點(diǎn)按從小到大的順序排列成一個(gè)數(shù)列，則該數(shù)列的 通項(xiàng)公式為(A . an =, n N* B . an = n( n 1, n N*C. an=n 1, n N* D . an=2n 2, n N*答案 C解析 當(dāng)x

7、W0寸，g(x = f(x x= 2x 1 x是減函數(shù)，只有一個(gè)零點(diǎn)a1 = 0;當(dāng) x>0 時(shí)，若 x = n, n N*,則 f(n = f(n 1 + 1 = = f(0 + n = n;若x不是整數(shù)，則 f(x= f(x 1 + 1 = = f(x x 1+ x + 1 ,其中x代表x的整數(shù)部分，由 f(x= x得 f(x x 1 = x x 1 ,其中1<x x 1<0,在(1,0沒(méi)有這樣的x. g(x= f(x x的零點(diǎn)按從小到大的順序?yàn)?,1,2,3,，通項(xiàng)an=n 1，故選C.3. 在直角坐標(biāo)系中，0是坐標(biāo)原點(diǎn)，P1(x1, y1 , P2(x2, y2是第一

8、象限的兩個(gè)點(diǎn)，若1,x1, x2,4依次成等差數(shù)列，而1, y1, y2,8依次成等比數(shù)列，則 0P1P2的面積是(A . 1 B . 2 C. 3 D . 4答案 A解析 由等差、等比數(shù)列的性質(zhì)，可求得 x1= 2, x2 = 3, y1 = 2, y2 = 4, P1(2,2, P2(3,4. S 0P1P2 = 1.二、填空題(每小題5分，共15分4. 已知數(shù)列an滿足：a1 = 1, an =n= 2,3,4,，設(shè)bn= a2n 1 + 1, n = 1,2,3,，則數(shù)列 bn的通項(xiàng)公式是 .答案 bn = 2n解析 由題意，得對(duì)于任意的正整數(shù)n, bn=a2n 1 + 1, bn+

9、1 = a2n+ 1,又 a2n + 1 = (2a+ 1 + 1 = 2(a2n 1 + 1= 2bn,/ bn+ 1 = 2bn,又 b1= a1+ 1 = 2, bn是首項(xiàng)為2，公比為2的等比數(shù)列，/ bn= 2n.5. 設(shè)數(shù)列an滿足al + 2a2 = 3,點(diǎn)Pn(n, an對(duì)任意的n N* ,都有PnPn+ 1 = (1,2，則數(shù)列an的前 n項(xiàng)禾廿Sn=.答案 n(n 解析/ PnPn+ 1 = OPn + 1 =(n +1, an + 1 (n, an= (1, an+ 1 an = (1,2, an+ 1 an= 2. an是公差為2的等差數(shù)列.由 a1 + 2a2= 3,得

10、 a1 =, Sn= + n(n 1X2= n(n .6. 若數(shù)列an滿足一=d(n N* , d為常數(shù)，則稱數(shù)列an為調(diào)和數(shù)列，已知數(shù)列為調(diào)和數(shù)列且 x1 + x2+-+ x20= 200，貝V x5 + x16=.答案 20解析 由題意知，若an為調(diào)和數(shù)列，則為等差數(shù)列，由為調(diào)和數(shù)列，可得數(shù)列xn為等差數(shù)列，由等差數(shù)列的性質(zhì)知，x5 + x16= x1 + x20= x2 + x19=x10+ x11 = = 20.三、解答題7. (13分已知數(shù)列an的前n項(xiàng)和Sn與通項(xiàng)an滿足Sn= an.(1求數(shù)列an的通項(xiàng)公式；(2 設(shè) f(x = log3x, bn = f (a1 + f(a2

11、+卜 f (a n, Tn = + + +，求 T2 012;(3 若 cn = an (an,求cn的前 n 項(xiàng)和 Un.解(1 當(dāng) n= 1 時(shí)，al =,當(dāng) n時(shí)，an = Sn Sn 1,又 Sn= an,所以 an = an 1,即數(shù)列an是首項(xiàng)為，公比為的等比數(shù)列，故 an= n.(2 由已知可得 f (an= log3n = n,貝H bn= 1 2 3 一 n =,故=2,又 Tn= 2=2,所以 T2 012=.(3由題意得cn= ( n ,故 Un= c1 + c2+ + cn則 Un= ,兩式相減可得Un =+ n n+1=+ n+ n n + 1,則 Un = + n+ n n +1.答案 2n 1解析 由于 Sn= 2n an,所以 Sn+ 1 = 2(n +1 an+ 1,后式減去前式，得 Sn+ 1 Sn= 2 an+ 1 + an，即卩 an+ 1 = an + 1,變形為 an+ 1 2= (an 2,則數(shù)列an 2是以 a1 2 為首項(xiàng)，為 公比的等比數(shù)列又a1= 2 a1,l卩a1 = 1.則 an 2 = ( 1n 1,所以 an= 2 n 1.7.已知等比數(shù)列中，各項(xiàng)都是正數(shù)，且a1, a3,2a2成等差數(shù)列，則的值為 .答案 3 + 2解析 設(shè)等比數(shù)列an的公比為q,/ a1, a3