1、双基限时练(十)1等差数列an中,a11,a3a514,其前n项和Sn100,则n()A9B10C11 D12解析a11,a3a52a16d14,d2,Snn2100.n10.答案B2设Sn是等差数列an的前n项和,若S735,则a4()A8 B7C6 D5解析S7735,a1a710,a45.答案D3设数列an是单调递增的等差数列,前三项的和为12,前三项的积为48,则它的首项是()A1 B2C4 D8解析依题意a1a32a2,a24.解得或an是递增数列,a12.答案B4若数列an为等差数列,公差为,且S100145,则a2a4a100的值为()A60 B85C. D其他值解析设a1a3a
2、99S1,则a2a4a100S150d.依题意,有S1S150d145.又d,S160.a2a4a100602585.答案B5记等差数列的前n项和为Sn,若S24,S420,则该数列的公差d等于()A2 B3C6 D7解析由题意,有a1a24,a1a2a3a420,a3a416.a12da22d16.4d12,d3.答案B6在小于100的自然数中,所有被7除余2的数之和为()A765 B665C763 D663解析被7除余2的自然数,构成等差数列,其首项a12,公差d7.最大的an93,由2(n1)793得n14.这些数的和为S14665.答案B7在数列an中,an4n,a1a2anan2bn
3、,(nN*),其中a,b为常数,则ab_.解析an4n,a1.又知an为等差数列,且d4,an2bna1a2ann42n2n.a2,b,ab1.答案18在等差数列an中,S46,S820,则S16_.解析S46,S8S4a5a6a7a820,a1a46,a5a814.a9a10a11a1222,a13a1630,S1672.答案729在数列an中,an1(nN*),且a5,则a3_.解析由an1,得,即,所以数列是公差为的等差数列,故2d221,即a31.答案110等差数列an的前n项和记为Sn,已知a1030,a2050.(1)求通项an;(2)若Sn242,求n.解(1)设an的首项为a1
4、,公差为d,则通项ana1(n1)d102n.(2)由Snna1d,Sn242,可得方程12n2242.解得n11或n22(舍去),n11.11已知an是一个等差数列,且a21,a55.(1)求an的通项an;(2)求an的前n项和Sn的最大值解(1)设an的公差为d,由已知条件解得ana1(n1)d2n5.(2)Snna1dn24n(n2)24,所以,当n2时,Sn取得最大值4.12已知等差数列an满足:a37,a5a726,an的前n项和为Sn.(1)求an及Sn;(2)令bn(nN),求数列bn的前n项和Tn.解(1)设等差数列an的首项为a1,公差为d.a37,a5a726,解得an32(n1)2n1,Sn3n2n22n.即an2n1,Snn22n.(2)由(1)知an2n1,bn.Tn,即数列bn的前n项和Tn.