等差数列前n项和性质第二课时

32,,:,,.nnnadSSnnd若数列为等差数列公差为为其前项和则数列也为等差数列且公差为性质2nSAnBnAnBnn32)2(2)1(:nbnann例如探知求索7152111111277576773122352315147522115752152515122222222,,(),:{},,,,,,,nnnnnnSSadadannnnanSaddadnnSaSnnnTadnn设数列的首项为成首项为公差为公差为的等差数列解法一7152719411521511972222224,,,,,(.:)(),,nnnnnnSSbbbnnnnnnnbbnbTn成等差数解列且公差为设由已知可得:法二例题示范2008200612007{}n2007,220082006naSSaSn例3、在等差数列中，前项和为S，且，求牛刀小试公差为k2d2324,,,.,.nnkkkkkSSSSaSnS若为等差数列为前项和也成等性则差数列质1221222122322122322232()(:)().:..kkkkkkkkkkkkkkkkkkkkkSaaaSSaaaakdakdakdSkdSSaaaSSkdSSSSSkd＋＋＋＋由，＝同理所以，，也成等差数列，并且公差为解探知求索例.设等差数列{an}的前n项和为Sn,若S3=9,S6=36,则S9=________81解法1：依题意知，S3=9，S6=36将它们代入公式解法2：由题意知，设则有解法3：S3，S6–S3，S9–S6，成等差数列a7+a8+a9=?例题示范1111011221101099109101001101091002110110100991121001025011100101001001000010010111101111111010010,,,,:::nnaadSadaddAABSAnBnABBSnnS由设解法一解法二解法三则111001010010010111210011101110011101110011010201010090110100101009010010902110221102109101002(),,,(),,,,,,,,,:,naaSSSSaaaaaaaaaaaSSSSSSSSdnTTSd成等差数列设公差为前项和为则解法四1111011101022111001102,,dTSd例题示范48910111211423021003(),,___.(){},,________.:nSSaaaaammm已知等差数列中已知等差数列的前项和为前项的和为则数列的前项的和为练习4484128128910111211131111355130312270221221002331313321022(),,,,:,,,,.()()()::,()()():mSSSSSSSaaaammmadmmmadmmmadmmmmSmadmad成等差数列即①解法一,②①得②法一法232333307010027030210:,,,,,,mmmmmmmmSSSSSSSS二成等差数列成等差数列牛刀小试2121(21)(21)nnnnnnSnaaTnbb2121nnST探知求索22)12(22)12(2))(12(2))(12(1211211212nnnnnnbnanbbnaanTS.两等差数列{an}、{bn}的前n项和分别是Sn和Tn,且71427nnSnTn求和.55abnnab556463ab146823nnanbn例题示范21215959792659312:nnnnaSbTaSbT解由1，等差数列{an}{bn}的前n项和分别为/,nnSS（1）已知4388ba，求/1515SS.（2）已知4332/nnSSnn，求88ba.2.两等差数列{an}、{bn}的前n项和分别是Sn和Tn,且22mnSmTnmnab，则牛刀小试