6.2等差數(shù)列及其前n項(xiàng)和,知識(shí)梳理,考點(diǎn)自測(cè),1.等差數(shù)列(1)定義:一般地,如果一個(gè)數(shù)列從起,每一項(xiàng)與它的前一項(xiàng)的等于,那么這個(gè)數(shù)列就叫做等差數(shù)列,這個(gè)常數(shù)叫做等差數(shù)列的,公差通常用字母d表示.數(shù)學(xué)語(yǔ)言表示為(nN*),d為常數(shù).(2)等差中項(xiàng):數(shù)列a,A,b成等差數(shù)列的充要條件是,其中A叫做a,b的.(3)等差數(shù)列的通項(xiàng)公式:an=,可推廣為an=am+(n-m)d.,第2項(xiàng),差,同一個(gè)常數(shù),公差,an+1-an=d,等差中項(xiàng),a1+(n-1)d,知識(shí)梳理,考點(diǎn)自測(cè),2.等差數(shù)列的通項(xiàng)公式及前n項(xiàng)和公式與函數(shù)的關(guān)系(1)an=a1+(n-1)d可化為an=dn+a1-d的形式.當(dāng)d0時(shí),an是關(guān)于n的一次函數(shù);當(dāng)d>0時(shí),數(shù)列為遞增數(shù)列;當(dāng)d0”是“S4+S6>2S5”的()A.充分不必要條件B.必要不充分條件C.充分必要條件D.既不充分也不必要條件,C,解析:因?yàn)樗許4+S6>2S510a1+21d>10a1+20dd>0,即“d>0”是“S4+S6>2S5”的充分必要條件,選C.,3.(2017遼寧撫順重點(diǎn)校一模,文2)在等差數(shù)列an中,a3+a6=11,a5+a8=39,則公差d為()A.-14B.-7C.7D.14,C,解析:a3+a6=11,a5+a8=39,則4d=28,解得d=7.故選C.,知識(shí)梳理,考點(diǎn)自測(cè),4.已知an為等差數(shù)列,Sn為其前n項(xiàng)和.若a1=6,a3+a5=0,則S6=.,6,解析:an是等差數(shù)列,a3+a5=2a4=0.a4=0.a4-a1=3d=-6.d=-2.S6=6a1+15d=66+15(-2)=6.,18162,考點(diǎn)一,考點(diǎn)二,考點(diǎn)三,考點(diǎn)四,等差數(shù)列中基本量的求解例1(1)(2017遼寧大連一模,文6)已知數(shù)列an滿足an+1-an=2,a1=-5,則|a1|+|a2|+|a6|=()A.9B.15C.18D.30(2)設(shè)等差數(shù)列an的前n項(xiàng)和為Sn,若Sm-1=-2,Sm=0,Sm+1=3,則m等于()A.3B.4C.5D.6,C,C,考點(diǎn)一,考點(diǎn)二,考點(diǎn)三,考點(diǎn)四,考點(diǎn)一,考點(diǎn)二,考點(diǎn)三,考點(diǎn)四,考點(diǎn)一,考點(diǎn)二,考點(diǎn)三,考點(diǎn)四,思考求等差數(shù)列基本量的一般方法是什么?解題心得1.等差數(shù)列運(yùn)算問題的一般求法是設(shè)出首項(xiàng)a1和公差d,然后由通項(xiàng)公式或前n項(xiàng)和公式轉(zhuǎn)化為方程(組)求解.2.等差數(shù)列的通項(xiàng)公式及前n項(xiàng)和公式共涉及五個(gè)量a1,an,d,n,Sn,已知其中三個(gè)就能求出另外兩個(gè),體現(xiàn)了用方程組解決問題的思想.3.減少運(yùn)算量的設(shè)元的技巧,若三個(gè)數(shù)成等差數(shù)列,可設(shè)這三個(gè)數(shù)分別為a-d,a,a+d;若四個(gè)數(shù)成等差數(shù)列,可設(shè)這四個(gè)數(shù)分別為a-3d,a-d,a+d,a+3d.,考點(diǎn)一,考點(diǎn)二,考點(diǎn)三,考點(diǎn)四,對(duì)點(diǎn)訓(xùn)練1(1)已知等差數(shù)列an前9項(xiàng)的和為27,a10=8,則a100=()A.100B.99C.98D.97(2)(2017福建廈門一模,文14)已知an是等差數(shù)列,其前n項(xiàng)和為Sn,a1+a3+a5=15,a2+a4+a6=0,則Sn的最大值為.,C,30,考點(diǎn)一,考點(diǎn)二,考點(diǎn)三,考點(diǎn)四,考點(diǎn)一,考點(diǎn)二,考點(diǎn)三,考點(diǎn)四,等差數(shù)列的判定與證明,考點(diǎn)一,考點(diǎn)二,考點(diǎn)三,考點(diǎn)四,思考判斷或證明一個(gè)數(shù)列為等差數(shù)列的基本方法有哪些?解題心得1.等差數(shù)列的四種判斷方法:(1)定義法:an+1-an=d(d是常數(shù))an是等差數(shù)列.(2)等差中項(xiàng)法:2an+1=an+an+2(nN*)an是等差數(shù)列.(3)通項(xiàng)公式:an=pn+q(p,q為常數(shù))an是等差數(shù)列.(4)前n項(xiàng)和公式:Sn=An2+Bn(A,B為常數(shù))an是等差數(shù)列.2.若證明一個(gè)數(shù)列不是等差數(shù)列,則只需證明存在連續(xù)三項(xiàng)不成等差數(shù)列即可.,考點(diǎn)一,考點(diǎn)二,考點(diǎn)三,考點(diǎn)四,對(duì)點(diǎn)訓(xùn)練2設(shè)數(shù)列an的前n項(xiàng)和為Sn,且Sn=2n-1.數(shù)列bn滿足b1=2,bn+1-2bn=8an.(1)求數(shù)列an的通項(xiàng)公式;(2)證明:數(shù)列為等差數(shù)列,并求bn的通項(xiàng)公式.,考點(diǎn)一,考點(diǎn)二,考點(diǎn)三,考點(diǎn)四,等差數(shù)列性質(zhì)的應(yīng)用(多考向)考向1等差數(shù)列項(xiàng)的性質(zhì)的應(yīng)用例3(1)(2017福建龍巖一模,文3)在等差數(shù)列an中,a3,a7是函數(shù)f(x)=x2-4x+3的兩個(gè)零點(diǎn),則an的前9項(xiàng)和等于()A.-18B.9C.18D.36(2)已知an是等差數(shù)列,Sn是其前n項(xiàng)和.若=-3,S5=10,則a9的值是.,C,20,解析:(1)等差數(shù)列an中,a3,a7是函數(shù)f(x)=x2-4x+3的兩個(gè)零點(diǎn),a3+a7=4,(2)由S5=10,得a3=2,因此2-2d+(2-d)2=-3,即d=3,故a9=2+36=20.,思考如何快捷地求出結(jié)果?,考點(diǎn)一,考點(diǎn)二,考點(diǎn)三,考點(diǎn)四,考向2等差數(shù)列前n項(xiàng)和的性質(zhì)的應(yīng)用例4在等差數(shù)列an中,前m項(xiàng)的和為30,前2m項(xiàng)的和為100,則前3m項(xiàng)的和為.,210,思考本例題應(yīng)用什么性質(zhì)求解比較簡(jiǎn)便?解題心得在等差數(shù)列an中,依據(jù)題意應(yīng)用其前n項(xiàng)和的性質(zhì)解題能比較簡(jiǎn)便地求出結(jié)果,常用的性質(zhì)有:在等差數(shù)列an中,數(shù)列Sm,也是等差數(shù)列.,考點(diǎn)一,考點(diǎn)二,考點(diǎn)三,考點(diǎn)四,A,5,考點(diǎn)一,考點(diǎn)二,考點(diǎn)三,考點(diǎn)四,考點(diǎn)一,考點(diǎn)二,考點(diǎn)三,考點(diǎn)四,等差數(shù)列前n項(xiàng)和的最值問題例5(2017北京海淀模擬)等差數(shù)列an中,設(shè)Sn為其前n項(xiàng)和,且a1>0,S3=S11,則當(dāng)n為多少時(shí),Sn最大?,考點(diǎn)一,考點(diǎn)二,考點(diǎn)三,考點(diǎn)四,解得6.5n7.5,故當(dāng)n=7時(shí),Sn最大.法四:由S3=S11,可得2a1+13d=0,即(a1+6d)+(a1+7d)=0,故a7+a8=0,又由a1>0,S3=S11可知d0,a8<0,所以當(dāng)n=7時(shí),Sn最大.,思考求等差數(shù)列前n項(xiàng)和的最值有哪些方法?,考點(diǎn)一,考點(diǎn)二,考點(diǎn)三,考點(diǎn)四,解題心得求等差數(shù)列前n項(xiàng)和Sn最值的兩種方法:(1)函數(shù)法:將等差數(shù)列的前n項(xiàng)和Sn=An2+Bn(A,B為常數(shù))看作二次函數(shù),根據(jù)二次函數(shù)的性質(zhì)求最值.,考點(diǎn)一,考點(diǎn)二,考點(diǎn)三,考點(diǎn)四,對(duì)點(diǎn)訓(xùn)練4等差數(shù)列an的前n項(xiàng)和為Sn,已知S10=0,S15=25,則nSn的最小值為多少?,考點(diǎn)一,考點(diǎn)二,考點(diǎn)三,考點(diǎn)四,1.等差數(shù)列的判斷方法(1)定義法;(2)等差中項(xiàng)法;(3)利用通項(xiàng)公式判斷;(4)利用前n項(xiàng)和公式判斷.2.公差不為0的等差數(shù)列的前n項(xiàng)和公式是n的二次函數(shù),且常數(shù)項(xiàng)為0.若某數(shù)列的前n項(xiàng)和公式是常數(shù)項(xiàng)不為0的二次函數(shù),則該數(shù)列不是等差數(shù)列,它從第2項(xiàng)起成等差數(shù)列.3.方程思想和化歸思想:在解有關(guān)等差數(shù)列的問題時(shí),可以先考慮把已知條件都化歸為a1和d等基本量的關(guān)系,再通過建立方程(組)求解.,考點(diǎn)一,考點(diǎn)二,考點(diǎn)三,考點(diǎn)四,1.當(dāng)公差d0時(shí),等差數(shù)列的通項(xiàng)公式是n的一次函數(shù);當(dāng)公差d=0時(shí),an為常數(shù).2.注意利用“an-an-1=d”時(shí)加上條件“n2”;否則,當(dāng)n=1時(shí),a0無(wú)定義.,