必修5 2.3.04 等差数列的前n项和的性质

等差数列前n项和的性质等差数列平均分组,各组之和仍为等差数列。232,,nmmmmmaSSSSS，如果为等差数列则也成等差数列。2mSmd，新的等差数列首项为公差为。等差数列前n项和的性质1：1、已知{an}是等差数列：(1)a1+a2+a3=5，a4+a5+a6=10,则a7+a8+a9=＿＿＿a19+a20+a21=_____(2)Sn=25,S2n=100,则S3n=＿＿＿＿＿1535225课堂练习。是等差数列，公差为，，，解：由已知可得数列32421620484dSSSSS12932)15(41620SSS所以1620201817SSaaa因为。求若公差项和为的前等差数列220191817412aaaaSdSnann，，，、}{。的值为129201817aaa课堂练习项，则共有若等差数列12nan21(1)(21)nnSna(2)1SnSn奇偶奇数项个数偶数项个数(3)nSSa奇偶中间项等差数列前n项和的性质2：1352124622(3)(...)(...)nnSSaaaaaaaa奇偶nnnnnaannaaanaan)1(2))(1(2)(222121项，则共有若等差数列12nan12121(21)()(21)2(1)(21)22nnnnnaanaSna等差数列前n项和的性质2：1352124622(...)(2)(...)nnSaaaaSaaaa奇偶121222()2(1)()(1)12nnnnnaanannaanan)...()...(264212531nnaaaaaaaaSS偶奇1221212)(2)(nnnnaaaanaan项，则共有若等差数列nan2中间两项之比偶奇1)2(nnaaSSndSS奇偶)1(等差数列前n项和的性质2：)...()...(264212531nnaaaaaaaaSS偶奇1221212)(2)(nnnnaaaanaan项，则共有若等差数列nan2中间两项之比偶奇1)2(nnaaSSndSS奇偶)1(等差数列前n项和的性质2：)...()...(125312642nnaaaaaaaaSS奇偶ndaaaaaaaann)(...)()()(1225634122nnannAB数列是公差为d的等差数列，则SnSAnBnnSn是等差数列，公差为A.nnnaan2ndnSS已知是公差为d的等差数列，为数列的前项是等差数列，则公差为和，.等差数列前n项和的性质3：求公差项和为的前等差数列3，，、2dSnann}{)())(1(10864297531aaaaaaaaaa864297531)2(aaaaaaaaa105)1(d答案：45)2(4433已知项数为奇数的等差数列，奇数项的和为，偶数项的和为，求该数列的项数项解：设数列共有12n)...()...(2264212531nnaaaaaaaaSS偶奇1)1(2))(1(2)(222121nnannaaanaannnnn343344项该数列共有7dan，求公差：偶数项之比为项中奇数项与前项和为的前等差数列322712,35412}{)...()...(264212531nnaaaaaaaaSS偶奇1221212)(2)(nnnnaaaanaan354)(6)(62)(127612112112aaaaaaS3227761aaaaSSnn偶奇532,2776daa在项数为2n项的等差数列中，各奇数项的和为75，偶数项的和为90，末项与首项的差为27，求n)...()...(125312642nnaaaaaaaaSS奇偶ndaaaaaaaann)(...)()()(12256341227)12(12dnaan157590ndSS奇偶3,5dn解得：的值求且中，公差已知等差数列10032199531n...60...,21aaaaaaaaad两等差数列前n项和与通项的关系若数列{an}与{bn}都是等差数列,且前n项的和分别为Sn和Tn,则nnab2121nnST1212112121(21)()2(21)()2nnnnnaaSnbbT(21)(21)nnnnnaanbbnnnnananaanS)12(22)12(2))(12(1211288,3213,,,bannTSTSnbannnnnn求且项和分别为的前等差数列3431521153151515158888TSbaba22111112)(2)8()(4)0,1(84nnnnnnnnnnaaaaaNSSaaan+1nn+1a解数列是a：a等差数列。，，、正项数列例2)2(814nnnaSa}{是等差数列。求证}{na)1(项和的最小值。的前，求数列若nbabnnn}{3021)2(211112(2)28Saa1（）由知a解：1154223102931,02215,15225nnnnnanbnbnnNbnbS15的前项为负，最小且S，，、正项数列例2)2(814nnnaSa}{是等差数列。求证}{na)1(项和的最小值。的前，求数列若nbabnnn}{3021)2(