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1、課時知能訓(xùn)練一、選擇題1(2020全國卷)設(shè)Sn為等差數(shù)列an的前n項(xiàng)和,若a11,公差d2,Sk2Sk24,則k()A8B7C6D5【解析】數(shù)列an是等差數(shù)列,a11,d2.an2n1,又Sk2Sk24,ak2ak12(k2)2(k1)24k424,k5.【答案】D2設(shè)等差數(shù)列an的前n項(xiàng)和為Sn.若a111,a4a66,則當(dāng)Sn取最小值時,n等于()A6 B7 C8 D9【解析】設(shè)等差數(shù)列的公差為d,由a4a6a1a911a96,得到a95,從而d2.所以Sn11nn(n1)n212n,因此當(dāng)Sn取得最小值時,n6.【答案】A3已知數(shù)列an,則a1a2a3a4a99a100()A4 800
2、 B4 900 C5 000 D5 100【解析】由題意得a1a2a3a4a99a1000224498981002(24698)1005 000.【答案】C4設(shè)等差數(shù)列an的前n項(xiàng)和為Sn,若S39,S636,則a7a8a9等于()A63 B45 C36 D27【解析】S3、S6S3,S9S6成等差數(shù)列,且S39,S636,S6S327,a7a8a9S318245.【答案】B5(2020惠州質(zhì)檢)已知數(shù)列an滿足a11,an1,則an等于()A. B.C. D.【解析】由題設(shè),得3,3.故是首項(xiàng)為1,公差為3的等差數(shù)列因此13(n1)3n2,an.【答案】C二、填空題6(2020湖南高考)設(shè)S
3、n是等差數(shù)列an(nN*)的前n項(xiàng)和,且a11,a47,則S5_.【解析】由a4a13d得d2,S551225.【答案】257等差數(shù)列an的前n項(xiàng)和為Sn,且6S55S35,則a4_.【解析】6S55S35,6(5a110d)5(3a13d)5,a13d,即a4.【答案】8各項(xiàng)均不為零的等差數(shù)列an中,若aan1an10(nN*,n2),則S2 012等于_【解析】an1an12anaan1an1a2an0,解得an2或an0(舍)S2 01222 0124 024.【答案】4 024三、解答題9在數(shù)列an中,a11,an12an2n,設(shè)bn,證明:數(shù)列bn是等差數(shù)列【證明】an12an2nb
4、n11bn1,bn1bn1,又b1a11,數(shù)列bn是首項(xiàng)為1,公差為1的等差數(shù)列10已知數(shù)列an中a18,a42,且滿足an2an2an1.(1)求數(shù)列an的通項(xiàng)公式;(2)設(shè)Sn是數(shù)列|an|的前n項(xiàng)和,求Sn.【解】(1)由2an1an2an可得an是等差數(shù)列,且公差d2.ana1(n1)d2n10.(2)令an0,得n5.當(dāng)n5時,an0;n6時,an0.當(dāng)n5時,Sn|a1|a2|an|a1a2ann29n當(dāng)n6時,Sn|a1|a2|an|a1a2a5(a6a7an)(a1a2an)2(a1a2a5)(n29n)2(5245)n29n40,Sn11(2020肇慶模擬)設(shè)a1,d為實(shí)數(shù),首項(xiàng)為a1,公差為d的等差數(shù)列an的前n項(xiàng)和為Sn,滿足S5S6150.(1)若S55,求S6及a1;(2)求d的取值范圍【解】(1)由題意知S63,a6S6S58.所以解得a17,所以S63,a17.(2)S5S6150,(5a110d)(6a115d)150,即2a9da110d210.由于關(guān)于a1的一元二次方程有解,所以81d28(10d21)d280,解得d2或d2.