再讀教材2025高考數(shù)學(xué)專項(xiàng)復(fù)習(xí)第九章　統(tǒng)計與成對數(shù)據(jù)的統(tǒng)計分析

再讀教材2025高考數(shù)學(xué)專項(xiàng)復(fù)習(xí)第九章統(tǒng)計與成對數(shù)據(jù)的統(tǒng)計分析第九章統(tǒng)計與成對數(shù)據(jù)的統(tǒng)計分析[知識網(wǎng)絡(luò)][命題方向]1．統(tǒng)計與成對數(shù)據(jù)的統(tǒng)計分析作為高考的必考內(nèi)容，在2025年的高考中預(yù)計仍會以“一小一大”的格局呈現(xiàn)．2．小題一般比較簡單，出現(xiàn)在選擇題或填空題中比較靠前的位置，命題角度主要有：統(tǒng)計數(shù)據(jù)的分析，多以統(tǒng)計圖表(折線圖或柱狀圖)的形式提供數(shù)據(jù)，進(jìn)行數(shù)據(jù)的特征分析，如均值、方差、最值點(diǎn)及趨勢分析等．3．解答題屬于中檔以上題目，題目涉及兩個以上的知識模塊，具有一定的綜合性．命題角度主要有三個方面：一是統(tǒng)計圖表與分布列的綜合；二是統(tǒng)計數(shù)據(jù)的數(shù)字特征與回歸分析、獨(dú)立性檢驗(yàn)等的綜合，此類問題計算量較大，注重數(shù)據(jù)的分析與應(yīng)用；三是統(tǒng)計圖表與函數(shù)內(nèi)容的結(jié)合，包括函數(shù)解析式的求解與應(yīng)用等，這有可能重新成為命題的熱點(diǎn)．探究1(人教A版必修第二冊P174)假設(shè)口袋中有紅色和白色共有1000個小球，除顏色外，小球的大小、質(zhì)地完全相同．你能通過抽樣調(diào)查的方法估計袋中紅球所占的比例嗎？____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________探究2(人教A版必修第二冊P179)小明想考察一下簡單隨機(jī)抽樣的估計效果．他從樹人中學(xué)醫(yī)務(wù)室得到了高一年級學(xué)生身高的所有數(shù)據(jù)，計算出整個年級學(xué)生的平均身高為165.0cm.然后，小明用簡單隨機(jī)抽樣的方法，從這些數(shù)據(jù)中抽取了樣本量為50和100的樣本各10個，分別計算出樣本平均數(shù)，如下表所示．從小明多次抽樣所得的結(jié)果中，你有什么發(fā)現(xiàn)？抽樣序號12345678910樣本量為50的平均數(shù)165.2162.8164.4164.4165.6164.8165.3164.7165.7165.0樣本量為100的平均數(shù)164.4165.0164.7164.9164.6164.9165.1165.2165.1165.2____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________探究3(人教A版必修第二冊P183)與考察簡單隨機(jī)抽樣估計效果類似，小明也想通過多次抽樣考察一下分層隨機(jī)抽樣的估計效果．他用比例分配的分層隨機(jī)抽樣方法，從高一年級的學(xué)生中抽取了10個樣本量為50的樣本，計算出樣本平均數(shù)如下表所示．與上一小節(jié)“探究”中相同樣本量的簡單隨機(jī)抽樣的結(jié)果比較，小明有了一個重要的發(fā)現(xiàn)．你是否也有所發(fā)現(xiàn)？抽樣序號12345678910男生樣本的平均數(shù)170.0170.7169.8171.7172.7171.9171.6170.6172.6170.9女生樣本的平均數(shù)162.2160.3159.7158.1161.1158.4159.7160.0160.6160.2總樣本的平均數(shù)165.8165.1164.3164.3166.4164.6165.2164.9166.1165.1____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________探究4(人教A版必修第二冊P197)分別以3和27為組數(shù)，對數(shù)據(jù)進(jìn)行等距分組，畫出100戶居民用戶月均用水量的頻率分布直方圖．觀察圖形，你發(fā)現(xiàn)不同的組數(shù)對于直方圖呈現(xiàn)數(shù)據(jù)分布規(guī)律有什么影響？(1)“組數(shù)為3”(2)“組數(shù)為27”______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________探究5(人教A版必修第二冊P206)平均數(shù)和中位數(shù)都描述了數(shù)據(jù)的集中趨勢，它們的大小關(guān)系和數(shù)據(jù)分布的形態(tài)有關(guān)．在圖中的三種分布形態(tài)中，平均數(shù)和中位數(shù)的大小存在什么關(guān)系？_________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________探究6(人教A版選擇性必修第三冊P94)在對人體的脂肪含量和年齡之間關(guān)系的研究中，科研人員獲得了一些年齡和脂肪含量的簡單隨機(jī)樣本數(shù)據(jù)，如表所示．表中每個編號下的年齡和脂肪含量數(shù)據(jù)都是對同一個體的觀測結(jié)果，它們構(gòu)成了成對數(shù)據(jù)．編號1234567年齡/歲23273941454950脂肪含量/%9.517.821.225.927.526.328.2編號891011121314年齡/歲53545657586061脂肪含量/%29.630.231.430.833.535.234.6根據(jù)以上數(shù)據(jù)，你能推斷人體的脂肪含量與年齡之間存在怎樣的關(guān)系嗎？_________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________典題1(人教A版必修第二冊P181T1)為了合理調(diào)配電力資源，某市欲了解全市50000戶居民的日用電量．若通過簡單隨機(jī)抽樣從中抽取了300戶進(jìn)行調(diào)查，得到其日用電量的平均數(shù)為5.5kW·h，則可以推測全市居民用戶日用電量的平均數(shù)()A．一定為5.5kW·h B．高于5.5kW·hC．低于5.5kW·h D．約為5.5kW·h_______________________________________________________________________________________________________________________________________________________________________________________________________________典題2(人教A版必修第二冊P184T1)數(shù)據(jù)x1，x2，…，xm的平均數(shù)為eq\o(x,\s\up6(－))，數(shù)據(jù)y1，y2，…，yn的平均數(shù)為eq\o(y,\s\up6(－))，證明：eq\f(\o(∑,\s\up6(m),\s\do4(i＝1))xi＋\o(∑,\s\up6(n),\s\do4(i＝1))yi,m＋n)＝eq\f(m,m＋n)eq\o(x,\s\up6(－))＋eq\f(n,m＋n)eq\o(y,\s\up6(－)).___________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________典題3(人教A版必修第二冊P189T6)數(shù)據(jù)x1，x2，…，xn的平均數(shù)為eq\o(x,\s\up6(－))，數(shù)據(jù)y1，y2，…，yn的平均數(shù)為eq\o(y,\s\up6(－))，a，b為常數(shù)．如果滿足y1＝ax1＋b，y2＝ax2＋b，…，yn＝axn＋b.證明：eq\o(y,\s\up6(－))＝aeq\o(x,\s\up6(－))＋b.________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________典題4(人教A版必修第二冊P216T4)數(shù)據(jù)x1，x2，…，xn的方差和標(biāo)準(zhǔn)差分別為seq\o\al(2,x)，sx，數(shù)據(jù)y1，y2，…，yn的方差和標(biāo)準(zhǔn)差分別為seq\o\al(2,y)，sy，若y1＝ax1＋b，y2＝ax2＋b，…，yn＝axn＋b成立，a，b為常數(shù)，證明：seq\o\al(2,y)＝a2seq\o\al(2,x)，sy＝|a|sx.______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________(多選)(2021·新高考Ⅰ卷)有一組樣本數(shù)據(jù)x1，x2，…，xn，由這組數(shù)據(jù)得到新樣本數(shù)據(jù)y1，y2，…，yn，其中yi＝xi＋c(i＝1，2，…，n)，c為非零常數(shù)，則()A．兩組樣本數(shù)據(jù)的樣本平均數(shù)相同B．兩組樣本數(shù)據(jù)的樣本中位數(shù)相同C．兩組樣本數(shù)據(jù)的樣本標(biāo)準(zhǔn)差相同D．兩組樣本數(shù)據(jù)的樣本極差相同點(diǎn)評本題考查平均數(shù)、中位數(shù)、標(biāo)準(zhǔn)差、極差的定義等基礎(chǔ)知識，基礎(chǔ)題與課本題相似度極高．典題5(人教A版必修第二冊P198T1)從某小區(qū)抽取100戶居民用戶進(jìn)行月用電量調(diào)查，發(fā)現(xiàn)他們的用電量都在50～350kW·h之間，進(jìn)行適當(dāng)分組后(每組為左閉右開的區(qū)間)，畫出頻率分布直方圖如圖所示．(1)直方圖中x的值為________；(2)在被調(diào)查的用戶中，用電量落在區(qū)間[100，250)內(nèi)的戶數(shù)為________．_______________________________________________________________________________________________________________________________________________________________________________________________________________(2022·新高考Ⅱ卷)在某地區(qū)進(jìn)行流行病學(xué)調(diào)查，隨機(jī)調(diào)查了100位某種疾病患者的年齡，得到如下的樣本數(shù)據(jù)的頻率分布直方圖：(1)估計該地區(qū)這種疾病患者的平均年齡(同一組中的數(shù)據(jù)用該組區(qū)間的中點(diǎn)值為代表)；(2)估計該地區(qū)一位這種疾病患者的年齡位于區(qū)間[20，70)的概率；(3)已知該地區(qū)這種疾病的患病率為0.1%，該地區(qū)年齡位于區(qū)間[40，50)的人口占該地區(qū)總?cè)丝诘?6%.從該地區(qū)中任選一人，若此人的年齡位于區(qū)間[40，50)，求此人患這種疾病的概率(以樣本數(shù)據(jù)中患者的年齡位于各區(qū)間的頻率作為患者的年齡位于該區(qū)間的概率，精確到0.0001.)_________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________點(diǎn)評本題考查了頻率分布直方圖，屬于中檔題．典題6(人教A版必修第二冊P203例2)根據(jù)9.1.2節(jié)問題3中女生的樣本數(shù)據(jù)，女生的原始數(shù)據(jù)為：163．0164.0161.0157.0162.0165.0158．0155.0164.0162.5154．0154.0164.0149.0159.0161.0170．0171.0155.0148.0172．0162.5158.0155.5157.0163.0172.0估計樹人中學(xué)高一年級女生的第25，50，75百分位數(shù)．______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________典題7(人教A版必修第二冊P204例3)根據(jù)如圖所示的頻率分布表或頻率分布直方圖，估計月均用水量的樣本數(shù)據(jù)的80%和95%分位數(shù)．分組頻數(shù)累計頻數(shù)頻率[1.2，4.2)230.23[4.2，7.2)320.32[7.2，10.2)130.13[10.2，13.2)90.09[13.2，16.2)90.09[16.2，19.2)50.05[19.2，22.2)30.03[22.2，25.2)40.04[25.2，28.2]20.02合計1001.00_________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________典題8(人教A版必修第二冊P208)借助居民月均用水量的頻率分布直方圖．如圖，你能估算出樣本的平均數(shù)、中位數(shù)和眾數(shù)嗎？___________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________典題9(人教A版必修第二冊P216T3)在去年的足球聯(lián)賽上，一隊(duì)每場比賽平均失球數(shù)是1.5，全年比賽失球個數(shù)的標(biāo)準(zhǔn)差為1.1；二隊(duì)每場比賽平均失球數(shù)是2.1，全年失球個數(shù)的標(biāo)準(zhǔn)差是0.4.你認(rèn)為下列說法中哪一種是正確的，為什么？(1)平均說來一隊(duì)比二隊(duì)防守技術(shù)好；(2)二隊(duì)比一隊(duì)技術(shù)水平更穩(wěn)定；(3)一隊(duì)有時表現(xiàn)很差，有時表現(xiàn)又非常好；(4)二隊(duì)很少不失球．_____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________典題10(人教A版必修第二冊P224T2)四名同學(xué)各擲骰子5次，分別記錄每次骰子出現(xiàn)的點(diǎn)數(shù)．根據(jù)四名同學(xué)的統(tǒng)計結(jié)果，可以判斷出一定沒有出現(xiàn)點(diǎn)數(shù)6的是()A．平均數(shù)3，中位數(shù)為2 B．中位數(shù)為3，眾數(shù)為2C．平均數(shù)為2，方差為2.4 D．中位數(shù)為3，方差為2.8_______________________________________________________________________________________________________________________________________________________________________________________________________________(2019·全國Ⅱ卷)演講比賽共有9位評委分別給出某選手的原始評分，評定該選手的成績時，從9個原始評分中去掉1個最高分、1個最低分，得到7個有效評分.7個有效評分與9個原始評分相比，不變的數(shù)字特征是()A．中位數(shù) B．平均數(shù)C．方差 D．極差點(diǎn)評本題考查數(shù)據(jù)的數(shù)字特征，關(guān)鍵是掌握數(shù)據(jù)的平均數(shù)，中位數(shù)，方差，極差的定義以及計算方法，與課本中習(xí)題考點(diǎn)一致，屬于基礎(chǔ)題．典題11(人教A版選擇性必修第三冊P101例1)在對人體的脂肪含量和年齡之間關(guān)系的研究中，科研人員獲得了一些年齡和脂肪含量的簡單隨機(jī)樣本數(shù)據(jù)，如表所示．表中每個編號下的年齡和脂肪含量數(shù)據(jù)都是對同一個體的觀測結(jié)果，它們構(gòu)成了成對數(shù)據(jù)．編號1234567年齡/歲23273941454950脂肪含量/%9.517.821.225.927.526.328.2編號891011121314年齡/歲53545657586061脂肪含量/%29.630.231.430.833.535.234.6根據(jù)上表中脂肪含量和年齡的樣本數(shù)據(jù)，推斷兩個變量是否線性相關(guān)，計算樣本相關(guān)系數(shù)，并推斷它們的相關(guān)程度．_________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________(2022·全國乙卷)某地經(jīng)過多年的環(huán)境治理，已將荒山改造成了綠水青山．為估計一林區(qū)某種樹木的總材積量，隨機(jī)選取了10棵這種樹木，測量每棵樹的根部橫截面積(單位：m2)和材積量(單位：m3)，得到如下數(shù)據(jù)：樣本號i根部橫截面積xi材積量yi10.040.2520.060.4030.040.2240.080.5450.080.5160.050.3470.050.3680.070.4690.070.42100.060.40總和0.63.9并計算得eq\o(∑,\s\up6(10),\s\do4(i＝1))xeq\o\al(2,i)＝0.038，eq\o(∑,\s\up6(10),\s\do4(i＝1))yeq\o\al(2,i)＝1.6158，eq\o(∑,\s\up6(10),\s\do4(i＝1))xiyi＝0.2474.(1)估計該林區(qū)這種樹木平均一棵的根部橫截面積與平均一棵的材積量；(2)求該林區(qū)這種樹木的根部橫截面積與材積量的樣本相關(guān)系數(shù)(精確到0.01)；(3)現(xiàn)測量了該林區(qū)所有這種樹木的根部橫截面積，并得到所有這種樹木的根部橫截面積總和為186m2.已知樹木的材積量與其根部橫截面積近似成正比．利用以上數(shù)據(jù)給出該林區(qū)這種樹木的總材積量的估計值．附：相關(guān)系數(shù)r＝eq\f(\o(∑,\s\up6(n),\s\do4(i＝1))（xi－\o(x,\s\up6(－))）（yi－\o(y,\s\up6(－))）,\r(\o(∑,\s\up6(n),\s\do4(i＝1))（xi－\o(x,\s\up6(－))）2\o(∑,\s\up6(n),\s\do4(i＝1))（yi－\o(y,\s\up6(－))）2))，eq\r(1.896)≈1.377.___________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________點(diǎn)評本題考查相關(guān)系數(shù)的求法，考查計算能力，與課本中例題相似度高．典題12(人教A版選擇性必修第三冊P132例3)某兒童醫(yī)院用甲、乙兩種療法治療小兒消化不良．采用有放回簡單隨機(jī)抽樣的方法對治療情況進(jìn)行檢查，得到了如下數(shù)據(jù)：抽到接受甲種療法的患兒67名，其中未治愈15名，治愈52名；抽到接受乙種療法的患兒69名，其中未治愈6名，治愈63名．試根據(jù)小概率值α＝0.005的獨(dú)立性檢驗(yàn)，分析乙種療法的效果是否比甲種療法好．___________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________(2022·全國甲卷)甲、乙兩城之間的長途客車均由A和B兩家公司運(yùn)營．為了解這兩家公司長途客車的運(yùn)行情況，隨機(jī)調(diào)查了甲、乙兩城之間的500個班次，得到下面列聯(lián)表：準(zhǔn)點(diǎn)班次數(shù)未準(zhǔn)點(diǎn)班次數(shù)A24020B21030(1)根據(jù)上表，分別估計這兩家公司甲、乙兩城之間的長途客車準(zhǔn)點(diǎn)的概率；(2)根據(jù)小概率α＝0.1的獨(dú)立性檢驗(yàn)，能否認(rèn)為甲、乙兩城之間的長途客車準(zhǔn)點(diǎn)與客車所屬公司有關(guān)？附：χ2＝eq\f(n（ad－bc）2,（a＋b）（c＋d）（a＋c）（b＋d）)，α0.1000.0500.010xα2.7063.8416.635___________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________點(diǎn)評本題考查了統(tǒng)計與概率中的獨(dú)立性檢驗(yàn)，屬于基礎(chǔ)題，且與課本例題命題角度類似．第六章數(shù)列[知識網(wǎng)絡(luò)][命題方向]1．本章高考試題難度以中檔為主，題型一般為一小(選擇題或填空題)一大(解答題)，總分值約為17分．(1)在高考試題中多以等差數(shù)列、等比數(shù)列的基本量運(yùn)算為載體，以數(shù)列遞推關(guān)系形式出現(xiàn)，考查數(shù)列求和及數(shù)列最值等綜合問題．(2)在處理等差、等比數(shù)列基本量運(yùn)算，遞推關(guān)系求通項(xiàng)，數(shù)列求和等問題時，常用公式法．(3)本章重點(diǎn)考查的學(xué)科核心素養(yǎng)為數(shù)學(xué)運(yùn)算和邏輯推理．2．考查內(nèi)容也較為穩(wěn)定，主要是以下幾個方面：(1)以等差、等比數(shù)列基本量的運(yùn)算為載體，考查等差數(shù)列、等比數(shù)列的概念、性質(zhì)、通項(xiàng)公式的求解與應(yīng)用；(2)考查數(shù)列求和的綜合問題，涉及數(shù)列的最值及解決方法；(3)考查數(shù)學(xué)文化、實(shí)際應(yīng)用為背景的數(shù)列問題．探究1(人教A版選擇性必修第二冊P14)觀察等差數(shù)列的通項(xiàng)公式，你認(rèn)為它與我們熟悉的哪一類函數(shù)有關(guān)？_______________________________________________________________________________________________________________________________________________________________________________________________________________探究2(人教A版選擇性必修第二冊P22)已知數(shù)列{an}的前n項(xiàng)和為Sn＝pn2＋qn＋r，其中p，q，r為常數(shù)，且p≠0.觀察數(shù)列{an}的特點(diǎn)，研究它是一個怎樣的數(shù)列，并證明你的結(jié)論．____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________探究3(人教A版選擇性必修第二冊P24)已知等差數(shù)列{an}的前n項(xiàng)和為Sn，若a1＝10，當(dāng)d＝－3.5時，Sn有最大值嗎？考慮更一般的等差數(shù)列前n項(xiàng)和的最大值問題．____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________探究4(人教A版選擇性必修第二冊P32)已知b＞0且b≠1，如果數(shù)列{an}是等差數(shù)列，那么數(shù)列{ban}是否一定是等比數(shù)列？如果數(shù)列{an}是各項(xiàng)均為正的等比數(shù)列，那么數(shù)列{logban}是否一定是等差數(shù)列？____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________典題1(人教A版選擇性必修第二冊P9習(xí)題4.1T7)已知函數(shù)f(x)＝eq\f(2x－1,2x)(x∈R)，設(shè)數(shù)列{an}的通項(xiàng)公式為an＝f(n)(n∈N*)．(1)求證an≥eq\f(1,2).(2){an}是遞增數(shù)列還是遞減數(shù)列？為什么？______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________典題2(人教A版選擇性必修第二冊P16例3)某公司購置了一臺價值為220萬元的設(shè)備，隨著設(shè)備在使用過程中老化，其價值會逐年減少．經(jīng)驗(yàn)表明，每經(jīng)過一年其價值就會減少d(d為正常數(shù))萬元．已知這臺設(shè)備的使用年限為10年，超過10年，它的價值將低于購進(jìn)價值的5%，設(shè)備將報廢．請確定d的取值范圍．______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________典題3(人教A版選擇性必修第二冊P17例5)已知數(shù)列{an}是等差數(shù)列，p，q，s，t∈N*，且p＋q＝s＋t.求證ap＋aq＝as＋at.___________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________典題4(人教A版選擇性必修第二冊P18練習(xí)T4)已知數(shù)列{an}，{bn}都是等差數(shù)列，公差分別為d1，d2，數(shù)列{cn}滿足cn＝an＋2bn.(1)數(shù)列{cn}是否是等差數(shù)列？若是，證明你的結(jié)論；若不是，請說明理由．(2)若{an}，{bn}的公差都等于2，a1＝b1＝1，求數(shù)列{cn}的通項(xiàng)公式．_________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________典題5(人教A版選擇性必修第二冊P18練習(xí)T5)已知一個無窮等差數(shù)列{an}的首項(xiàng)為a1，公差為d.(1)將數(shù)列中的前m項(xiàng)去掉，其余各項(xiàng)組成一個新的數(shù)列，這個新數(shù)列是等差數(shù)列嗎？如果是，它的首項(xiàng)和公差分別是多少？(2)取出數(shù)列中的所有奇數(shù)項(xiàng)，組成一個新的數(shù)列，這個新數(shù)列是等差數(shù)列嗎？如果是，它的首項(xiàng)和公差分別是多少？(3)取出數(shù)列中所有序號為7的倍數(shù)的項(xiàng)，組成一個新的數(shù)列，它是等差數(shù)列嗎？你能根據(jù)得到的結(jié)論作出一個猜想嗎？_________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________典題6(人教A版選擇性必修第二冊P21例6)已知數(shù)列{an}是等差數(shù)列．(1)若a1＝7，a50＝101，求S50；(2)若a1＝2，a2＝eq\f(5,2)，求S10；(3)若a1＝eq\f(1,2)，d＝－eq\f(1,6)，Sn＝－5，求n._________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________典題7(人教A版選擇性必修第二冊P21例7)已知一個等差數(shù)列{an}前10項(xiàng)的和是310，前20項(xiàng)的和是1220.由這些條件能確定這個等差數(shù)列的首項(xiàng)和公差嗎？____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________典題8(人教A版選擇性必修第二冊P23練習(xí)T4)在等差數(shù)列{an}中，若S15＝5(a2＋a6＋ak)，求k.____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________(2023·全國甲卷)記Sn為等差數(shù)列{an}的前n項(xiàng)和．若a2＋a6＝10，a4a8＝45，則S5＝()A．25 B．22C．20 D．15點(diǎn)評本題與教材習(xí)題都是考查等差數(shù)列中的基本量的計算．典題9(人教A版選擇性必修第二冊P23練習(xí)T5)已知一個等差數(shù)列的項(xiàng)數(shù)為奇數(shù)，其中所有奇數(shù)項(xiàng)的和為290，所有偶數(shù)項(xiàng)的和為261.求此數(shù)列中間一項(xiàng)的值以及項(xiàng)數(shù)．_________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________典題10(人教A版選擇性必修第二冊P23例9)已知等差數(shù)列{an}的前n項(xiàng)和為Sn，若a1＝10，公差d＝－2，則Sn是否存在最大值？若存在，求Sn的最大值及取得最大值時n的值；若不存在，請說明理由．_________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________(2022·全國甲卷)記Sn為數(shù)列{an}的前n項(xiàng)和，已知eq\f(2Sn,n)＋n＝2an＋1.(1)證明：{an}是等差數(shù)列；(2)若a4，a7，a9成等比數(shù)列，求Sn的最小值．_________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________點(diǎn)評本題與教材例題都是考查前n項(xiàng)和Sn的最值問題．典題11(人教A版選擇性必修第二冊P24練習(xí)T5)已知數(shù)列{an}的通項(xiàng)公式為an＝eq\f(n－2,2n－15)，前n項(xiàng)和為Sn，求Sn取得最小值時n的值．______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________典題12(人教A版選擇性必修第二冊P25習(xí)題4.2T2)已知{an}為等差數(shù)列，a1＋a3＋a5＝105，a2＋a4＋a6＝99.求a20.______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________(2020·全國Ⅰ卷)設(shè){an}是等比數(shù)列，且a1＋a2＋a3＝1，a2＋a3＋a4＝2，則a6＋a7＋a8＝()A．12 B．24C．30 D．32點(diǎn)評本題在所給條件的形式上與教材習(xí)題很類似，只是把等差數(shù)列改為了等比數(shù)列，解題的難點(diǎn)和關(guān)鍵點(diǎn)分別是求公差、公比．典題13(人教A版選擇性必修第二冊P25習(xí)題4.2T7)已知Sn是等差數(shù)列{an}的前n項(xiàng)和．(1)證明eq\b\lc\{\rc\}(\a\vs4\al\co1(\f(Sn,n)))是等差數(shù)列；(2)設(shè)Tn為數(shù)列eq\b\lc\{\rc\}(\a\vs4\al\co1(\f(Sn,n)))的前n項(xiàng)和，若S4＝12，S8＝40，求Tn._________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________典題14(人教A版選擇性必修第二冊P25習(xí)題4.2T8)已知兩個等差數(shù)列2，6，10，…，190及2，8，14，…，200，將這兩個等差數(shù)列的公共項(xiàng)按從小到大的順序組成一個新數(shù)列．求這個新數(shù)列的各項(xiàng)之和．_________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________(2020·新高考全國Ⅰ卷)將數(shù)列{2n－1}與{3n－2}的公共項(xiàng)從小到大排列得到數(shù)列{an}，則{an}的前n項(xiàng)和為__________．點(diǎn)評本題和教材習(xí)題考查角度完全相同，由于不用判斷求和數(shù)列的項(xiàng)數(shù)，所以其難度要小于教材習(xí)題，解決此類問題的關(guān)鍵是清楚新組合的數(shù)列的公差是兩個數(shù)列公差的最小公倍數(shù)．典題15(人教A版選項(xiàng)性必修第二冊P25習(xí)題4.2T10)已知等差數(shù)列{an}的公差為d，求證eq\f(am－an,m－n)＝d.你能從直線的斜率角度來解釋這個結(jié)果嗎？______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________典題16(人教A版選擇性必修第二冊P26習(xí)題4.2T12)如圖的形狀出現(xiàn)在南宋數(shù)學(xué)家楊輝所著的《詳解九章算法·商功》中，后人稱為“三角垛”．“三角垛”的最上層有1個球，第二層有3個球，第三層有6個球……設(shè)各層球數(shù)構(gòu)成一個數(shù)列{an}．(1)寫出數(shù)列{an}的一個遞推公式；(2)根據(jù)(1)中的遞推公式，寫出數(shù)列{an}的一個通項(xiàng)公式．_________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________典題17(人教A版選擇性必修第二冊P30例2)已知等比數(shù)列{an}的公比為q，試用{an}的第m項(xiàng)am表示an.______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________典題18(人教A版選擇性必修第二冊P31例4)用10000元購買某個理財產(chǎn)品一年．(1)若以月利率0.400%的復(fù)利計息，12個月能獲得多少利息(精確到1元)?(2)若以季度復(fù)利計息，存4個季度，則當(dāng)每季度利率為多少時，按季結(jié)算的利息不少于按月結(jié)算的利息(精確到10－5)?______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________典題19(人教A版選擇性必修第二冊P34練習(xí)T5)已知數(shù)列{an}的通項(xiàng)公式為an＝eq\f(n3,3n)，求使an取得最大值時的n的值．___________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________典題20(人教A版選擇性必修第二冊P35例7)已知數(shù)列{an}是等比數(shù)列．(1)若a1＝eq\f(1,2)，q＝eq\f(1,2)，求S8；(2)若a1＝27，a9＝eq\f(1,243)，q＜0，求S8；(3)若a1＝8，q＝eq\f(1,2)，Sn＝eq\f(31,2)，求n.___________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________典題21(人教A版選擇性必修第二冊P36例8)已知等比數(shù)列{an}的首項(xiàng)為－1，前n項(xiàng)和為Sn.若eq\f(S10,S5)＝eq\f(31,32)，求公比q.________________________________________________________________________________________________________________________________________________________________________________________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