《高中数学北师大版必修5配套练习:1.2等差数列 第2课时》由会员分享,可在线阅读,更多相关《高中数学北师大版必修5配套练习:1.2等差数列 第2课时(5页珍藏版)》请在金锄头文库上搜索。
1、 第一章2 第2课时一、选择题1已知等差数列an中,a35,a59,则a7()A11B12C13D14答案C解析设公差为d,a5a32d,2d4,又a7a52d9413.2在等差数列an中,a3a4a5a6a7450,则a2a8()A45B75C180D300答案C解析由a3a7a4a62a5,得a3a7a4a6a55a5450,a590.a2a82a5180.3下列命题中正确的是()A若a,b,c成等差数列,则a2,b2,c2成等差数列B若a,b,c成等差数列,则log2a,log2b,log2c成等差数列C若a,b,c成等差数列,则a2,b2,c2成等差数列D若a,b,c成等差数列,则2a
2、,2b,2c成等差数列答案C解析a,b,c成等差数列,2bac,2b4ac4,即2(b2)(a2)(c2),a2,b2,c2成等差数列4已知等差数列an中,a7a916,a41,则a12等于()A15B30C31D64答案A解析a7a92a816,故a88.在等差数列an中,a4,a8,a12成等差数列,所以a122a8a416115.5已知等差数列an满足a1a2a3a1010,则有()Aa1a1010Ba2a1000,a36,a72.,故a110,d2,an2n12.10四个数成等差数列,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,
3、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1等差数列an中,a1a4a739,a2a5a833,则a3a6a9的值为()A30B27C24D21答案B解析解法一:设b1a1a4a739,b2a2a5a833,b3a3a6a9,an成等差数列,b1,b2,b3成等差数列,a3a6a9b3b2(b2b1)2b2b127.解法二:设等差数列an的公差为d,则a2a5a8a1a4a73d,33393
4、d,3d6,a3a6a9a2a5a83d33627.2设数列an,bn都是等差数列,且a125,b175,a2b2100,则a37b37等于()A0B37C100D37答案C解析a1b1100,a2b2100,(a2a1)(b2b1)0,设等差数列an,bn的公差分别为d1,d2,则d1d20.a37b37a136d1b136d2a1b136(d1d2)a1b1100.3数列an中,a22,a60且数列是等差数列,则a4等于()ABCD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,a4.4等差数列an中,a2a5a89,那么关于x的
5、方程:x2(a4a6)x100()A无实根B有两个相等实根C有两个不等实根D不能确定有无实根答案A解析a4a6a2a82a5,即3a59,a53,方程为x26x100,无实数解二、填空题5已知an为等差数列,a1a3a5105,a2a4a699,则a20_.答案1解析a1a3a5105,即3a3105a335,同理a433,da4a32a20a4(204)d1.6等差数列an中,公差为,且a1a3a5a9960,则a2a4a6a100_.答案85解析由等差数列的定义知a2a4a6a100a1a3a5a9950d602585.三、解答题7在等差数列an中,已知a2a5a89,a3a5a721,求数列的通项公式解析a2a5a89,a3a5a721,又a2a8a3a72a5,a3a72a56,即a53.a3a77,由、解得a31,a77,或a37,a71,a31,d2或a37,d2.由ana3(n3)d,得an2n7或an2n13.8已知5个数成等差数列,它们的和为5,平方和为,求这5个数解析设这五个数依次为a2d,ad,a,ad,a2d,由题意,得,解得,.故这五个数为,1,或,1,.