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高中数学 9.2 等差数列第2课时同步练习 湘教版必修41等差数列an中,a1a3a5,则cos a3()A B C D2在等差数列an中,a5a6a7a8a9450,则a3a11的值为()A45 B75 C180 D3003等差数列an中,a1a2a50200,a51a52a1002 700,则公差d等于()A1 B1 C5 D504等差数列an中,a13a8a15120,则2a9a10()A24 B22 C20 D85已知数列an满足:a11,an0,an21(nN*),那么使an5成立的n的最大值为()A4 B5 C24 D256已知等差数列an,bn的公差分别为2和3,且bnN*,则数列abn是()A等差数列且公差为5 B等差数列且公差为6C等差数列且公差为8 D等差数列且公差为97在等差数列an中,a3,a8是方程x23x50的两个根,则a1a10_.8已知数列an中,a11,an1anan1an,则数列an的通项公式为an_.9已知四个数成等差数列,它们的和为26,中间两项的积为40,求这四个数10在数列an中,a11,an12an2n.设,证明bn是等差数列,并求数列an的通项公式参考答案1. 答案:D解析:因为an是等差数列,所以a1a3a5(a1a5)a33a3,因此,故cos a3.2. 答案:C解析:a5a6a7a8a95a7450,于是a790,故a3a112a7180.3. 答案:B解析:由a1a2a5050,得50a50(12349)d200.由a51a52a10050,得50a50(1250)d2 700.得2 500d2 500,d1,故选B.4. 答案:A解析:由a13a8a15120得5a8120,所以a824,于是2a9a10a8a10a10a824.5. 答案:C解析:由a11,an0,an21(nN*)可得a n2n,即an,要使an5,则n25,故n的最大值是24.6. 答案:B解析:依题意有abna1(bn1)22bna122b12(n1)3a126na12b18,故abn1abn6,即数列abn是等差数列且公差为6,故选B.7. 答案:3解析:依题意得a3a83,于是a1a10a3a83.8. 答案:解析:由an1anan1an得,所以是公差为1的等差数列,又因为,所以1(n1)n,所以an.9. 答案:解:设这四个数为a3d,ad,ad,a3d.依题意有解得所以这四个数分别为2,5,8,11或11,8,5,2.10. 答案:解:由已知an12an2n得bn11bn1,又b1a11,bn是首项为1,公差为1的等差数列,因此,ann2n1.
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