www.76684.com:14.第八讲 等差数列与等比数列(一)(教师版)

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模块三 数列、复数、推理与证明 第八讲 等差数列与等比数列 第八讲 等差数列与等比数列(一) 等差数列 考点一 等差数列基本运算 定 通 义 项 an-an-1=d(n≥2) an= a1+(n-1)d = am+(n-m)d (m,n∈N*,且 n>m) 2A=a+b a,A,b 成等差 1.已知等差数列{an}中,a1=1,a3=-3. (1)求数列{an}的通项公式; (2)若数列{an}的前 k 项和 Sk=-35,求 k 的值. 解:(1)设等差数列{an}的公差为 d,则 an=a1+(n-1)d. 由 a1=1,a3=-3,可得 1+2d=-3,解得 d=-2.从而 an=1+(n-1)×(-2)=3-2n. (2)由(1)可知 an=3-2n,所以 Sn= =2n-n2. 由 Sk=-35,可得 2k-k2=-35,即 k2-2k-35=0,解得 k=7 或 k=-5.又 k∈N*,故 k=7. 2. (全国Ⅰ)已知等差数列{an}的前 n 项和 Sn 满足 S3=0,S5=-5.求{an}的通项公式 解:设{an}的公差为 d,则 Sn=na1+ d. 由已知可得 解得 a1=1,d=-1.故{an}的通项公式为 an=2-n. 3.已知等差数列{an}满足:a3=7,a5+a7=26,{an}的前 n 项和为 Sn,求 an 及 Sn; 解:设等差数列{an}的公差为 d, 因为 a3=7,a5+a7=26,所以有 解得 所以 an=3+2(n-1)=2n+1, Sn=3n+ ×2=n2+2n. 4.在等差数列{an}中,a1+a2=4,a7+a8=28,则数列的通项公式 an 为(C) A.2n B.2n+1 C.2n-1 D.2n+2 解析:由已知得:(a7+a8)-(a1+a2)=12d=24,∴d=2.∴a1+a1+d=4.∴a1=1. ∴an=a1+(n-1)d=1+(n-1)×2=2n-1. *高考是比知识,比能力,比心理,比信心,比体力* 1 模块三 数列、复数、推理与证明 第八讲 等差数列与等比数列 考点二 等差数列的性质 性 中 质 k+l=m+n 项 p+q=2m ak+al=am+an a p +aq ? 2am 5.在等差数列{an}中,a2+a6= π ,则 sin =(D) A. B. C.- D.- 解析:在等差数列{an}中,a2+a6= π =2a4,则 sin 6.在等差数列{an}中,已知 a4+a8=16,则 a2+a10=(B) A.12 B.16 C.20 D.24 =- ,选 D. 解析:由等差数列的性质知,a2+a10=a4+a8=16,故选 B. 7. (广东)在等差数列{an}中,已知 a3+a8=10,则 3a5+a7= 20 . 解析:因为 a3+a8=a5+a6=a4+a7=10.所以 3a5+a7=a5+2a5+a7=a5+a4+a6+a7=2×10=20. 考点三 等差数列求和公式应用 求 和 Sn= n(a1 ? an ) =na1+ n(n-1) d 2 2 Sn,S2n-Sn,S3n-S2n 成 2(S2n-Sn)=Sn+( S3n-S2n) 8. 等差数列{an}的前 n 项和为 Sn,若 a2=1,a3=3,则 S4 等于(C) A.12 B.10 C.8 D.6 解析:由题意 d=2,a1=-1,∴S4=4a1+ ×2=8.故选 C. 49 9. 设 Sn 是等差数列{an}的前 n 项和,已知 a2=3,S11=121,则 S7= 解析:由 S11= . =49. =11a6=121,则 a6=11.又 a2=3,则 a2+a6=a1+a7=14,∴S7= 10. 已知等差数列{an}的前 13 项之和为 ,则 tan(a6+a7+a8)等于(C) A. B. C.-1 D.1 解析:∵{an}的前 13 项之和为 =13a7,∴a7= ,则 tan(a6+a7+a8)=tan3a7=tan =-1,选 C. *高考是比知识,比能力,比心理,比信心,比体力* 2 模块三 数列、复数、推理与证明 第八讲 等差数列与等比数列 11. 等差数列{an}的前 n 项和为 Sn,若 S17 为一确定常数,则下列各式也为确定常数的是(C) A.a2+a15 B.a2·a15 C.a2+a9+a16 D.a2·a9·a16 解:因为 S17 为确定常数,则 a1+a17 为确定常数,又 a1+a17=a2+a16=2a9,所以 a2+a16+a9 为确定常数. ※注: S11= 11a6 S17=11a9 S11=11a6 能得到什么结论? ) 12. 设等差数列 {an } 的前 n 项和为 Sn ,若 S3 ? 9 , S6 ? 36 ,则 a7 ? a8 ? a9 ? ( A.63 B.45 C.36 D.27 解:由等差数列性质知 S3、S6-S3、S9-S6 成等差数列,即 9,27,S 成等差,所以 S=45,选 B. 13.已知在等差数列{an}中,a1=31,Sn 是它的前 n 项和,S10=S22. (1)求 Sn; (2)这个数列前多少项的和最大,并求出这个最大值. 解:(1)∵S10=a1+a2+…+a10,S22=a1+a2+…+a22, 又 S10=S22,∴a11+a12+…+a22=0, =0,即 a11+a22=2a1+31d=0,又 a1=31,∴d=-2, ∴Sn=na1+ d=31n-n(n-1)=32n-n2. (2)由(1)中可知 Sn=32n-n2=-(n-16)2+256,∴当 n=16 时,Sn 有最大值,Sn 的最大值是 256. 等差数列巩固练习 1.已知等差数列 ?an ? 中, a2 ? 7, a4 ? 15 ,则前10项的和 S10

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