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2022年高中数学 2.2第2课时 等差数列的性质练习 新人教A版必修5一、选择题1等差数列an中,a6a916,a41,则a11()A64 B30 C31 D15答案D解析解法一:,a11a110d15.解法二:69411,a4a11a6a916,a1115.2如果等差数列an中,a3a4a512,那么a1a2a7()A14 B21 C28 D35答案C解析a3a4a53a412,a44.又a1a2a77a428.3已知等差数列an满足a1a2a3a1010,则有()Aa1a1010 Ba2a1000Ca3a1000 Da510答案D解析由题设a1a2a3a101101a510,a510.4已知an为等差数列,a1a3a5105,a2a4a699,则a20等于()A1 B1 C3 D7答案B解析an是等差数列,a1a3a53a3105,a335,a2a4a63a499,a433,da4a32,a20a416d33321.5(xx陕西省质量监测)已知数列an满足a115,且3an13an2.若akak10,则正整数k()A24 B23 C22 D21答案B解析由3an13an2得an1an,所以数列an为首项a115,公差d的等差数列,所以an15(n1)n,则由akak10,ak10,a240,d3.则a11a12a133a123(a210d)105.二、填空题7等差数列an中,已知a2a3a10a1136,则a5a8_.答案18分析利用等差数列的性质求解,或整体考虑问题,求出2a111d的值解析解法1:根据题意,有(a1d)(a12d)(a19d)(a110d)36,4a122d36,则2a111d18.a5a8(a14d)(a17d)2a111d18.解法2:根据等差数列性质,可得a5a8a3a10a2a1136218.8已知等差数列an中,a3、a15是方程x26x10的两根,则a7a8a9a10a11_.答案15解析a3a156,又a7a11a8a102a9a3a15,a7a8a9a10a11(2)(a3a15)615.三、解答题9已知等差数列an的公差d0,且a3a712,a4a64,求an的通项公式解析由等差数列的性质,得a3a7a4a64,又a3a712,a3、a7是方程x24x120的两根又d0,a36,a72.a7a34d8,d2.ana3(n3)d62(n3)2n12.10四个数成等差数列,其平方和为94,第一个数与第四个数的积比第二个数与第三个数的积少18,求此四个数解析设四个数为a3d,ad,ad,a3d,据题意得,(a3d)2(ad)2(ad)2(a3d)2942a210d247.又(a3d)(a3d)(ad)(ad)188d218d代入得a,故所求四数为8,5,2,1或1,2,5,8或1,2,5,8或8,5,2,1.一、选择题11设数列an,bn都是等差数列,且a125,b175,a2b2100,那么数列anbn的第37项为()A0 B37 C100 D37答案C解析数列an,bn都是等差数列,anbn也是等差数列又a1b1100,a2b2100,anbn的公差为0,数列anbn的第37项为100.12数列an中,a22,a60且数列是等差数列,则a4等于()A B C D答案A解析令bn,则b2,b61,由条件知bn是等差数列,b6b2(62)d4d,d,b4b22d2,b4,a4.13(xx北京理,6)设an是等差数列下列结论中正确的是()A若a1a20,则a2a30B若a1a30,则a1a20C若0a1a2,则a2D若a10,则(a2a1)(a2a3)0答案C解析考查等差数列通项公式;作差比较法先分析四个答案,A举一反例a12,a21,则a34,a1a20,而a2a30,A错误;B举同样反例a12,a21,a34,a1a30,B错误;下面针对C进行研究,an是等差数列,若0a10,设公差为d,则d0,数列各项均为正,由于aa1a3(a1d)2a1(a12d)a2a1dd2a2a1dd20,则aa1a3a2,选C14下列命题中正确的个数是()(1)若a,b,c成等差数列,则a2,b2,c2一定成等差数列;(2)若a,b,c成等差数列,则2a,2b,2c可能成等差数列;(3)若a,b,c成等差数列,则ka2,kb2,kc2一定成等差数列;(4)若a,b,c成等差数列,则,可能成等差数列A4个 B3个 C2个 D1个答案B解析对于(1)取a1,b2,c3a21,b24,c29,(1)错对于(2),abc2a2b2c,(2)正确;对于(3),a,b,c成等差数列,ac2b.(ka2)(kc2)k(ac)42(kb2),(3)正确;对于(4),abc0,(4)正确,综上选B点评等差数列的性质(1)等差数列的项的对称性在有穷等差数列中,与首末两项“等距离”的两项之和等于首项与末项的和即a1ana2an1a3an2(2)若an、bn分别是公差为d,d的等差数列,则有数列结论can公差为d的等差数列(c为任一常数)can公差为cd的等差数列(c为任一常数)panqbn公差为pdqd的等差数列(p,q)为常数(3)an的公差为d,则d0an为递增数列;d4),则,解得a10,三边长分别为6,10,14.所以SABC61015.三、解答题17在ABC中,三边a、b、c成等差数列,、也成等差数列,求证ABC为正三角形证明2,平方得ac24b,又ac2b,b,故()20,abc.故ABC为正三角形18设数列an是等差数列,bn()an又b1b2b3,b1b2b3,求通项an.解析b1b2b3,又bn()an,()a1()a2()a3.()a1a2a3,a1a2a33,又an成等差数列a21,a1a32,b1b3,b1b3,或,即或,an2n3或an2n5.
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