新高考數(shù)學(xué)一輪復(fù)習(xí)知識點總結(jié)與題型精練專題16 等差數(shù)列及其前n項和(原卷版)

專題16等差數(shù)列及其前n項和【考綱要求】1、理解等差數(shù)列的定義，會推導(dǎo)等差數(shù)列的通項公式，能運用等差數(shù)列的通項公式解決一些簡單的問題.2、掌握等差中項的概念，深化認(rèn)識并能運用，掌握等差數(shù)列前n項和公式及其獲取思路.3、經(jīng)歷公式的推導(dǎo)過程，體驗從特殊到一般的研究方法，學(xué)會觀察、歸納、反思.4、熟練掌握等差數(shù)列的五個量a1，d，n，an，Sn的關(guān)系，能夠由其中三個求另外兩個．【思維導(dǎo)圖】一、等差數(shù)列的概念【考點總結(jié)】1、數(shù)列前n項和的概念如果一個數(shù)列從第2項起，每一項與它的前一項的差都等于同一個常數(shù)，那么這個數(shù)列就叫做等差數(shù)列，這個常數(shù)叫做等差數(shù)列的公差，公差通常用字母d表示．等差數(shù)列{an}的概念可用符號表示為an＋1－an＝d(n∈N*)．[化解疑難]1．“從第2項起”是指第1項前面沒有項，無法與后續(xù)條件中“與前一項的差”相吻合．2．“每一項與它的前一項的差”這一運算要求是指“相鄰且后項減去前項”，強調(diào)了：①作差的順序；②這兩項必須相鄰．3．定義中的“同一常數(shù)”是指全部的后項減去前一項都等于同一個常數(shù)，否則這個數(shù)列不能稱為等差數(shù)列.2、等差中項如果三個數(shù)a，A，b成等差數(shù)列，那么A叫做a與b的等差中項．這三個數(shù)滿足的關(guān)系式是A＝eq\f(a＋b,2).[化解疑難]1．A是a與b的等差中項，則A＝eq\f(a＋b,2)或2A＝a＋b，即兩個數(shù)的等差中項有且只有一個．2．當(dāng)2A＝a＋b時，A是a與b的等差中項.3、等差數(shù)列的通項公式已知等差數(shù)列{an}的首項為a1，公差為d遞推公式通項公式an－an－1＝d(n≥2)an＝a1＋(n－1)d(n∈N*)[化解疑難]由等差數(shù)列的通項公式an＝a1＋(n－1)d可得an＝dn＋(a1－d)，如果設(shè)p＝d，q＝a1－d，那么an＝pn＋q，其中p，q是常數(shù)．當(dāng)p≠0時，an是關(guān)于n的一次函數(shù)；當(dāng)p＝0時，an＝q，等差數(shù)列為常數(shù)列．二、等差數(shù)列的前n項和【考點總結(jié)】1、數(shù)列前n項和的概念把a1＋a2＋…＋an叫數(shù)列{an}的前n項和，記做Sn．則a1＋a2＋a3＋…＋an－1＝Sn－1(n≥2)．思考由Sn與Sn－1的表達式可以得出an＝eq\b\lc\{(\a\vs4\al\co1(Sn－Sn－1（n≥2），,S1（n＝1）.))2、等差數(shù)列前n項和公式1．公式1：若{an}是等差數(shù)列，則Sn可以用首項a1和末項an表示為Sn＝eq\f(n（a1＋an）,2)．2．公式2：若首項為a1，公差為d，則Sn可以表示為Sn＝na1＋eq\f(1,2)n(n－1)d．3．推導(dǎo)方法：倒序相加法過程：Sn＝a1＋a2＋…＋an，Sn＝an＋an－1＋…＋a1，∵a1＋an＝a2＋an－1＝…＝an＋a1，∴2Sn＝n(a1＋an)，∴Sn＝eq\f(n（a1＋an）,2).4．從函數(shù)角度認(rèn)識等差數(shù)列的前n項和公式(1)公式的變形Sn＝na1＋eq\f(n（n－1）d,2)＝eq\f(d,2)n2＋(a1－eq\f(d,2))n.(2)從函數(shù)角度認(rèn)識公式①當(dāng)d≠0時，Sn是項數(shù)n的二次函數(shù)，且不含常數(shù)項；②當(dāng)d＝0時，Sn＝na1，不是項數(shù)n的二次函數(shù)．3、等差數(shù)列前n項和的性質(zhì)1．若數(shù)列{an}是公差為d的等差數(shù)列，Sn為其前n項和，則數(shù)列eq\b\lc\{\rc\}(\a\vs4\al\co1(\f(Sn,n)))也是等差數(shù)列，且公差為eq\f(d,2).2．若Sm，S2m，S3m分別為等差數(shù)列{an}的前m項，前2m項，前3m項的和，則Sm，S2m－Sm，S3m－S2m也成等差數(shù)列，公差為m2d．3．設(shè)兩個等差數(shù)列{an}，{bn}的前n項和分別為Sn，Tn，則eq\f(an,bn)＝eq\f(S2n－1,T2n－1).4．若等差數(shù)列的項數(shù)為2n，則S2n＝n(an＋an＋1)，S偶－S奇＝nd，eq\f(S偶,S奇)＝eq\f(an＋1,an).5．若等差數(shù)列的項數(shù)為2n＋1，則S2n＋1＝(2n＋1)an＋1，S偶－S奇＝－an＋1，eq\f(S偶,S奇)＝eq\f(n,n＋1).【題型匯編】題型一：等差數(shù)列及其通項公式題型二：等差數(shù)列的性質(zhì)題型三：等差數(shù)列的前n項和題型四：等差數(shù)列的前n項和的函數(shù)特性【題型講解】題型一：等差數(shù)列及其通項公式一、單選題1．（2022·江西九江·三模（文））等差數(shù)列SKIPIF1<0中，若SKIPIF1<0，則SKIPIF1<0（

）A．16 B．18 C．20 D．222．（2022·四川成都·三模（文））在等差數(shù)列SKIPIF1<0中，已知SKIPIF1<0，SKIPIF1<0，則數(shù)列SKIPIF1<0的公差為（

）A．SKIPIF1<0 B．0 C．1 D．23．（2022·山西大附中三模（理））已知等差數(shù)列SKIPIF1<0的各項均為正數(shù)，其前n項和為SKIPIF1<0，且滿足SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0（

）A．28 B．30 C．32 D．354．（2022·陜西漢中·二模（理））已知等差數(shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則等差數(shù)列SKIPIF1<0的公差是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<05．（2022·廣西柳州·三模（文））記SKIPIF1<0為等差數(shù)列SKIPIF1<0的前SKIPIF1<0項和，若SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<06．（2022·寧夏·平羅中學(xué)三模（文））設(shè)等差數(shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0，則當(dāng)SKIPIF1<0取最小值時，SKIPIF1<0的值為（

）A．8 B．7 C．6 D．97．（2022·山西太原·一模（文））設(shè)SKIPIF1<0為等差數(shù)列SKIPIF1<0的前SKIPIF1<0項和，若SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0（

）A．26 B．27 C．28 D．298．（2022·四川雅安·二模）設(shè)等差數(shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0取最小值時，SKIPIF1<0的值為（

）A．19 B．20 C．21 D．20或219．（2022·四川成都·二模（理））已知數(shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0．若SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<010．（2022·江蘇·金陵中學(xué)二模）設(shè)SKIPIF1<0是公差SKIPIF1<0的等差數(shù)列，如果SKIPIF1<0，那么SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0題型二：等差數(shù)列的性質(zhì)一、單選題1．（2022·黑龍江·哈爾濱三中模擬預(yù)測（理））已知等差數(shù)列SKIPIF1<0的前n項和為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0（

）A．-110 B．-115 C．110 D．1152．（2022·北京東城·三模）在公差不為零的等差數(shù)列SKIPIF1<0中，若SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03．（2022·安徽淮南·二模（理））已知等差數(shù)列SKIPIF1<0的前n項和為SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0（

）A．8 B．12 C．14 D．204．（2022·安徽滁州·二模（文））已知SKIPIF1<0是公差不為零的等差數(shù)列，若SKIPIF1<0，則SKIPIF1<0（

）A．7 B．8 C．9 D．105．（2022·四川·成都七中二模（文））已知數(shù)列SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0（）A．6 B．7 C．8 D．96．（2022·廣東·潮州市瓷都中學(xué)三模）已知等差數(shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0（

）A．2020 B．1021 C．1010 D．10027．（2022·江西·二模（文））己知等差數(shù)列SKIPIF1<0的前n項和是SKIPIF1<0，若公差SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<08．（2022·河南許昌·三模（文））設(shè)等差數(shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<09．（2022·山西太原·二模（理））等差數(shù)列SKIPIF1<0的前n項和為SKIPIF1<0，若SKIPIF1<0則公差SKIPIF1<0（

）A．1 B．2 C．－1 D．－210．（2022·安徽省含山中學(xué)三模（文））已知等差數(shù)列SKIPIF1<0的前n項和為SKIPIF1<0．若SKIPIF1<0，則SKIPIF1<0（

）A．60 B．50 C．30 D．20二、多選題1．（2022·重慶·二模）設(shè)等差數(shù)列SKIPIF1<0前SKIPIF1<0項和為SKIPIF1<0，公差SKIPIF1<0，若SKIPIF1<0，則下列結(jié)論中正確的有（

）A．SKIPIF1<0 B．當(dāng)SKIPIF1<0時，SKIPIF1<0取得最小值C．SKIPIF1<0 D．當(dāng)SKIPIF1<0時，SKIPIF1<0的最小值為292．（2022·江蘇南京·二模）已知SKIPIF1<0是等差數(shù)列SKIPIF1<0的前SKIPIF1<0項和，且SKIPIF1<0，則下列說法正確的是（

）A．SKIPIF1<0中的最大項為SKIPIF1<0 B．?dāng)?shù)列SKIPIF1<0的公差SKIPIF1<0C．SKIPIF1<0 D．當(dāng)且僅當(dāng)SKIPIF1<0時，SKIPIF1<0題型三：等差數(shù)列的前n項和一、單選題1．（2022·遼寧沈陽·一模）已知等差數(shù)列SKIPIF1<0的公差為2，且SKIPIF1<0，SKIPIF1<0，SKIPIF1<0成等比數(shù)列，則SKIPIF1<0的前n項和SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．（2022·河南·一模（文））已知數(shù)列SKIPIF1<0為等差數(shù)列，首項SKIPIF1<0，公差SKIPIF1<0，前n項和SKIPIF1<0，則SKIPIF1<0（

）A．8 B．9 C．10 D．113．（2022·寧夏中衛(wèi)·三模（理））已知數(shù)列SKIPIF1<0滿足點SKIPIF1<0在直線SKIPIF1<0上，則數(shù)列SKIPIF1<0的前SKIPIF1<0項和SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<04．（2022·湖南省臨澧縣第一中學(xué)二模）設(shè)SKIPIF1<0為等差數(shù)列SKIPIF1<0的前SKIPIF1<0項和，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0A．-6 B．-4 C．-2 D．25．（2022·江西師大附中三模（理））等差數(shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0，滿足：SKIPIF1<0，則SKIPIF1<0（

）A．72 B．75 C．60 D．1006．（2022·內(nèi)蒙古呼和浩特·一模（理））已知在等差數(shù)列SKIPIF1<0中，SKIPIF1<0，則SKIPIF1<0（

）A．30 B．39 C．42 D．787．（2022·安徽合肥·二模（文））設(shè)等差數(shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的值為（

）A．10 B．12 C．13 D．148．（2022·重慶·二模）等差數(shù)列SKIPIF1<0的公差為2，前SKIPIF1<0項和為SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0的最大值為（

）A．3 B．6 C．9 D．129．（2022·內(nèi)蒙古呼和浩特·一模（文））記SKIPIF1<0為等差數(shù)列SKIPIF1<0的前n項和．若SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的公差為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<010．（2022·浙江杭州·二模）設(shè)等差數(shù)列SKIPIF1<0的前n項和為SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0（）A．12 B．15 C．18 D．21二、多選題1．（2022·河北滄州·二模）已知數(shù)列SKIPIF1<0滿足SKIPIF1<0，記SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<02．（2022·廣東惠州·二模）已知SKIPIF1<0為等差數(shù)列，其前SKIPIF1<0項和SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0，則（

）A．公差SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．當(dāng)且僅當(dāng)SKIPIF1<0時SKIPIF1<0題型四：等差數(shù)列的前n項和的函數(shù)特性一、單選題1．（2022·寧夏·平羅中學(xué)三模（文））設(shè)等差數(shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0，則當(dāng)SKIPIF1<0取最小值時，SKIPIF1<0的值為（

）A．8 B．7 C．6 D．92．（2022·四川雅安·二模）設(shè)等差數(shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0取最小值時，SKIPIF1<0的值為（

）A．19 B．20 C．21 D．20或213．（2022·重慶·二模）等差數(shù)列SKIPIF1<0的公差為2，前SKIPIF1<0項和為SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0的最大值為（

）A．3 B．6 C．9 D．124．（2022·河南許昌·三模（文））已知SKIPIF1<0是等差數(shù)列SKIPIF1<0的前n項和，若對任意的SKIPIF1<0，均有SKIPIF1<0．成立，則SKIPIF1<0的最小值為（

）A．2 B．SKIPIF1<0 C．3 D．SKIPIF1<05．（2022·北京·潞河中學(xué)三模）已知SKIPIF1<0是等差數(shù)列，SKIPIF1<0是其前SKIPIF1<0項和.則“SKIPIF1<0”是“對于任意SKIPIF1<0且SKIPIF1<0，SKIPIF1<0”