等差數(shù)列看一看【知識回憶】1．數(shù)列：按照一定順序排列著一列數(shù)．2．數(shù)列分類：遞增數(shù)列、遞減數(shù)列、常數(shù)列、擺動數(shù)列3．數(shù)列遞推公式：表示任一項SKIPIF1<0與它前一項SKIPIF1<0〔或前幾項〕間關系公式．4．對任意數(shù)列{}前項和與通項關系：5．等差數(shù)列定義：如果一個數(shù)列從第2項起，每一項與它前一項差等于同一個常數(shù)，那么這個數(shù)列稱為等差數(shù)列，這個常數(shù)稱為等差數(shù)列公差．符號表示:SKIPIF1<0。6．等差中項：由三個數(shù)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0組成等差數(shù)列可以看成最簡單等差數(shù)列，那么SKIPIF1<0稱為SKIPIF1<0與SKIPIF1<0等差中項．假設SKIPIF1<0，那么稱SKIPIF1<0為SKIPIF1<0與SKIPIF1<0等差中項．7．等差數(shù)列通項公式：假設等差數(shù)列SKIPIF1<0首項是SKIPIF1<0，公差是SKIPIF1<0，那么SKIPIF1<0．8．等差數(shù)列性質(zhì)：假設SKIPIF1<0是等差數(shù)列，且SKIPIF1<0〔SKIPIF1<0、SKIPIF1<0、SKIPIF1<0、SKIPIF1<0〕，那么SKIPIF1<0；假設SKIPIF1<0是等差數(shù)列，且SKIPIF1<0〔SKIPIF1<0、SKIPIF1<0、SKIPIF1<0〕，那么SKIPIF1<0．9．等差數(shù)列前SKIPIF1<0項和公式：=1\*GB3①SKIPIF1<0；=2\*GB3②SKIPIF1<0．想一想如何判定數(shù)列是不是等差數(shù)列？練一練1．在等差數(shù)列SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0，那么公差SKIPIF1<0等于〔〕A．B．0C．D．2．在等差數(shù)列SKIPIF1<0中，SKIPIF1<0那么該數(shù)列前SKIPIF1<0項和是〔〕A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<03．Sn是等差數(shù)列{an}前n項和，假設SKIPIF1<0QUOTEQUOTE，那么SKIPIF1<0QUOTEQUOTE=〔〕A．3B．5C．D．4．等差數(shù)列SKIPIF1<0前SKIPIF1<0項和為SKIPIF1<0，且滿足SKIPIF1<0，那么數(shù)列SKIPIF1<0公差是〔〕A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<05．數(shù)列SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0，假設SKIPIF1<0是等差數(shù)列，那么SKIPIF1<0等于〔〕A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<06．等差數(shù)列SKIPIF1<0前SKIPIF1<0項和為SKIPIF1<0，且滿足SKIPIF1<0,SKIPIF1<0，那么使SKIPIF1<0取最大值時SKIPIF1<0值為A．8B．10C．9或10D．8或97．如果在等差數(shù)列SKIPIF1<0中，SKIPIF1<0，那么SKIPIF1<0．8．在等差數(shù)列中，，那么．9．等差數(shù)列前項和為，假設，SKIPIF1<0，那么公差等于．10．等差數(shù)列SKIPIF1<0前SKIPIF1<0項和為SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0．〔1〕求數(shù)列SKIPIF1<0通項公式；〔2〕設SKIPIF1<0，求數(shù)列SKIPIF1<0中最小項．11．設SKIPIF1<0為等差數(shù)列SKIPIF1<0前SKIPIF1<0項和，且SKIPIF1<0．〔1〕求公差SKIPIF1<0值；〔2〕SKIPIF1<0，求所有滿足條件SKIPIF1<0值．12．等差數(shù)列中，．〔1〕求數(shù)列通項公式及前項和表達式；〔2〕記數(shù)列前項和為，求值．【等差數(shù)列】等差數(shù)列①anan1d(n2,d為常數(shù))②2anan1an1(n2)③anknb(n,k為常數(shù)1．A【解析】2a13d5，因為a14，代入d1an，應選試題分析：a2a3A.考點：等差數(shù)列2．B【解析】試題分析：由3a3a52a7a10a1324,有6a46a10a4a104,所以該數(shù)列前13項和13(a4a10)13426,應選B.22考點：1.等差數(shù)列性質(zhì)；2.等差數(shù)列前n項和公式.3．A【解析】試題分析：由題意得，設等差數(shù)列an公差為d，由a39a1，那么a12d9a1d4a1，所以S55a110d45a13，應選A．S33a13d15a1考點：等差數(shù)列通項及前n項和公式應用．4．D【解析】an)ann(a1Sna13S212，試題分析：因為Sn，所以，由2S3a1a32n2S3S2a1a2a3a2得2，得2，即aa4，所以數(shù)32222