新高考數(shù)學(xué)一輪復(fù)習(xí)4.1 等差數(shù)列（精講）（基礎(chǔ)版）（原卷版）

4.1等差數(shù)列（精講）（基礎(chǔ)版）思維導(dǎo)圖思維導(dǎo)圖考點(diǎn)呈現(xiàn)考點(diǎn)呈現(xiàn)例題剖析例題剖析考點(diǎn)一等差數(shù)列基本量的計(jì)算【例1】（2022·福建三明）已知等差數(shù)列{SKIPIF1<0}的前n項(xiàng)和為SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0＝（

）A．6 B．10 C．12 D．201.1.方程思想：等差數(shù)列的基本量為首項(xiàng)a1和公差d，通常利用已知條件及通項(xiàng)公式或前n項(xiàng)和公式列方程(組)求解，等差數(shù)列中包含a1，d，n，an，Sn五個(gè)量，可“知三求二”．2.整體思想：當(dāng)所給條件只有一個(gè)時(shí)，可將已知和所求都用a1，d表示，尋求兩者間的聯(lián)系，整體代換即可求解．3.利用性質(zhì)：運(yùn)用等差數(shù)列性質(zhì)可以化繁為簡(jiǎn)、優(yōu)化解題過(guò)程．溫馨提示【一隅三反】1．（2022·陜西漢中）已知等差數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)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<02．（2022·內(nèi)蒙古呼和浩特）已知在等差數(shù)列SKIPIF1<0中，SKIPIF1<0，則SKIPIF1<0（

）A．30 B．39 C．42 D．783．（2022·陜西·西安工業(yè)大學(xué)附中）設(shè)等差數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0（

）A．20 B．23 C．24 D．28考點(diǎn)二等差中項(xiàng)【例2-1】（2022·北京通州·一模）設(shè)等差數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0（

）A．60 B．70 C．120 D．140【例2-2】（2022·浙江杭州·二模）設(shè)等差數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0（）A．12 B．15 C．18 D．21【例2-3】（2022·安徽滁州）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．10【一隅三反】1．（2022·河北石家莊·二模）等差數(shù)列SKIPIF1<0的前n項(xiàng)和記為SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0（

）A．3033 B．4044 C．6066 D．80882．（2022·河南平頂山）已知SKIPIF1<0為正項(xiàng)等差數(shù)列SKIPIF1<0的前n項(xiàng)和，若SKIPIF1<0，則SKIPIF1<0（

）A．22 B．20 C．16 D．113．（2022·全國(guó)·高三專題練習(xí)）已知數(shù)列SKIPIF1<0滿足SKIPIF1<0且SKIPIF1<0，則SKIPIF1<0（

）A．-3 B．3 C．SKIPIF1<0 D．SKIPIF1<0考點(diǎn)三前n項(xiàng)和的性質(zhì)【例3-1】（2022·北京石景山）記SKIPIF1<0為等差數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和，若SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0（

）A．36 B．45 C．63 D．75【例3-2】（1）（2022·江西·臨川一中）已知數(shù)列SKIPIF1<0和SKIPIF1<0都是等差數(shù)列，且其前n項(xiàng)和分別為SKIPIF1<0和SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0（2）（2022·四川師范大學(xué)附屬中學(xué)二模（理））設(shè)等差數(shù)列SKIPIF1<0，SKIPIF1<0的前n項(xiàng)和分別是SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．3【例3-3】（2022·全國(guó)·高三專題練習(xí)）等差數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0，若SKIPIF1<0且SKIPIF1<0，則(

)A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【例3-4】（1）（2022·內(nèi)蒙古赤峰）已知等差數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0取最大值時(shí)正整數(shù)n的值為（

）A．9 B．10 C．11 D．12（2）（2022·重慶·二模）（多選）設(shè)等差數(shù)列SKIPIF1<0前SKIPIF1<0項(xiàng)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(shí)，SKIPIF1<0取得最小值C．SKIPIF1<0 D．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的最小值為291.等差數(shù)列前1.等差數(shù)列前n項(xiàng)和Sn最值的兩種方法(1)函數(shù)法：利用等差數(shù)列前n項(xiàng)和的函數(shù)表達(dá)式Sn＝an2＋bn，通過(guò)配方或借助圖象求二次函數(shù)最值的方法求解．(2)鄰項(xiàng)變號(hào)法：①當(dāng)a1>0，d<0時(shí)，滿足eq\b\lc\{\rc\(\a\vs4\al\co1(am≥0，,am＋1≤0))的項(xiàng)數(shù)m使得Sn取得最大值為Sm；②當(dāng)a1<0，d>0時(shí)，滿足eq\b\lc\{\rc\(\a\vs4\al\co1(am≤0，,am＋1≥0))的項(xiàng)數(shù)m使得Sn取得最小值為Sm.2.在等差數(shù)列中，Sn，S2n－Sn，S3n－S2n，…仍成等差數(shù)列；eq\b\lc\{\rc\}(\a\vs4\al\co1(\f(Sn,n)))也成等差數(shù)列溫馨提示【一隅三反】1．（2022·全國(guó)·高三專題練習(xí)）設(shè)等差數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0等于（

）A．-3 B．-12 C．-21 D．-302．（2022·全國(guó)·高三）若等差數(shù)列SKIPIF1<0和SKIPIF1<0的前n項(xiàng)的和分別是SKIPIF1<0和SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03．（2022·全國(guó)·高三專題練習(xí)）設(shè)等差數(shù)列SKIPIF1<0與等差數(shù)列SKIPIF1<0的前n項(xiàng)和分別為SKIPIF1<0，SKIPIF1<0，若對(duì)任意自然數(shù)n都有SKIPIF1<0，則SKIPIF1<0的值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<04．（2022·全國(guó)·高三專題練習(xí)）在等差數(shù)列SKIPIF1<0中，SKIPIF1<0，其前SKIPIF1<0項(xiàng)和為SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0等于（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<05．（2022·全國(guó)·高三專題練習(xí)）（多選）等差數(shù)列{an}的前n項(xiàng)和記為Sn，若a1>0，S10＝S20，則（

）A．d<0B．a(chǎn)16<0C．Sn≤S15D．當(dāng)且僅當(dāng)n≥32時(shí)，Sn<06．（2022·浙江省浦江中學(xué)高三期末）設(shè)等差數(shù)列SKIPIF1<0的公差為d，其前n項(xiàng)和為SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，則使得SKIPIF1<0的正整數(shù)n的最小值為（

）A．16 B．17 C．18 D．19考點(diǎn)四等差數(shù)列定義及其運(yùn)用【例4-1】（2022·全國(guó)·高三專題練習(xí)）（多選）下列數(shù)列是等差數(shù)列的是（

）A．0，0，0，0，0，… B．1，l，111，111l，…C．－5，－3，－1，1，3，… D．1，2，3，5，8，…【例4-2】（2022·全國(guó)·高三專題練習(xí)）在數(shù)列SKIPIF1<0中，有SKIPIF1<0,證明：數(shù)列SKIPIF1<0為等差數(shù)列，并求其通項(xiàng)公式.【例4-3】（2022·四川·瀘縣五中模擬預(yù)測(cè)（理））下列選項(xiàng)中，為“數(shù)列SKIPIF1<0是等差數(shù)列”的一個(gè)充分不必要條件的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．通項(xiàng)公式SKIPIF1<0 D．SKIPIF1<0【例4-4】（2022·全國(guó)·高三專題練習(xí)）已知不全相等的實(shí)數(shù)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0成等比數(shù)列，則一定不可能是等差數(shù)列的為（

）A．SKIPIF1<0，SKIPIF1<0，SKIPIF1<0 B．SKIPIF1<0，SKIPIF1<0，SKIPIF1<0 C．SKIPIF1<0，SKIPIF1<0，SKIPIF1<0 D．SKIPIF1<0，SKIPIF1<0，SKIPIF1<0等差數(shù)列的判定與證明的方法等差數(shù)列的判定與證明的方法方法解讀適合題型定義法若an－an－1(n≥2，n∈N*)為同一常數(shù)?{an}是等差數(shù)列解答題中證明問(wèn)題等差中項(xiàng)法2an＝an＋1＋an－1(n≥2，n∈N*)成立?{an}是等差數(shù)列通項(xiàng)公式法an＝pn＋q(p，q為常數(shù))對(duì)任意的正整數(shù)n都成立?{an}是等差數(shù)列選擇、填空題中的判定問(wèn)題前n項(xiàng)和公式驗(yàn)證Sn＝An2＋Bn(A，B是常數(shù))對(duì)任意的正整數(shù)n都成立?{an}是等差數(shù)列溫馨提示【一隅三反】1．（2022·全國(guó)·課時(shí)練習(xí)）（多選）若SKIPIF1<0是等差數(shù)列，則下列數(shù)列為等差數(shù)列的有（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．（2022·全國(guó)·高二課時(shí)練習(xí)）（多選）在數(shù)列SKIPIF1<0中，SKIPIF1<0，且對(duì)任意大于SKIPIF1<0的正整數(shù)SKIPIF1<0，點(diǎn)SKIPIF1<0在直線SKIPIF1<0上，則（

）A．?dāng)?shù)列SKIPIF1<0是等差數(shù)列B．?dāng)?shù)列SKIPIF1<0是等差數(shù)列C．?dāng)?shù)列SKIPIF1<0的通項(xiàng)公式為SKIPIF1<0D．?dāng)?shù)列SKIPIF1<0的通項(xiàng)公式為SKIPIF1<03．（2022·全國(guó)·課時(shí)練習(xí)）（多選）下列數(shù)列中是等差數(shù)列的是（

）A．SKIPIF1<0，a，SKIPIF1<0B．2，4，6，8，…，SKIPIF1<0，SKIPIF1<0C．SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0D．SKIPIF1<04．（2022·全國(guó)·高三專題練習(xí)）已知數(shù)列SKIPIF1<0中，SKIPIF1<0，當(dāng)n≥2時(shí)，SKIPIF1<0.求證:數(shù)列SKIPIF1<0是等差數(shù)列.考點(diǎn)五等差數(shù)列的實(shí)際應(yīng)用【例5】（2022·海南·文昌中學(xué)高三階段練習(xí)）《周髀算經(jīng)》是中國(guó)古代天文學(xué)與數(shù)學(xué)著作，其中有關(guān)于24節(jié)氣的描述，將一年分為24個(gè)節(jié)氣，如圖所示，已知晷長(zhǎng)指太陽(yáng)照射物體影子的長(zhǎng)度，相鄰兩個(gè)節(jié)氣的晷長(zhǎng)變化量相同（即每?jī)蓚€(gè)相鄰節(jié)氣晷長(zhǎng)增加或減小量相同，其中冬至晷長(zhǎng)最長(zhǎng)，夏至晷長(zhǎng)最短，從夏至到冬至晷長(zhǎng)逐漸變大，從冬至到夏