新高考數(shù)學(xué)一輪復(fù)習(xí)講練測(cè)第6章第02講 等差數(shù)列及其前n項(xiàng)和（練習(xí)）（原卷版）

第02講等差數(shù)列及其前n項(xiàng)和（模擬精練+真題演練）1．（2023·河南鄭州·統(tǒng)考模擬預(yù)測(cè)）在等差數(shù)列SKIPIF1<0中，已知SKIPIF1<0，且SKIPIF1<0，則當(dāng)SKIPIF1<0取最大值時(shí)，SKIPIF1<0（

）A．10 B．11 C．12或13 D．132．（2023·江蘇南通·統(tǒng)考模擬預(yù)測(cè)）現(xiàn)有茶壺九只，容積從小到大成等差數(shù)列，最小的三只茶壺容積之和為0.5升，最大的三只茶壺容積之和為2.5升，則從小到大第5只茶壺的容積為（

）A．0.25升 B．0.5升 C．1升 D．1.5升3．（2023·河南洛陽(yáng)·模擬預(yù)測(cè)）已知等差數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0（

）A．54 B．71 C．80 D．814．（2023·河南·校聯(lián)考模擬預(yù)測(cè)）已知數(shù)列SKIPIF1<0是等差數(shù)列，其前SKIPIF1<0項(xiàng)和為SKIPIF1<0，則SKIPIF1<0等于（

）A．63 B．SKIPIF1<0 C．45 D．SKIPIF1<05．（2023·北京海淀·校考三模）已知等差數(shù)列SKIPIF1<0的公差為SKIPIF1<0，數(shù)列SKIPIF1<0滿足SKIPIF1<0，則“SKIPIF1<0”是“SKIPIF1<0為遞減數(shù)列”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6．（2023·河南鄭州·統(tǒng)考模擬預(yù)測(cè)）公差不為零的等差數(shù)列SKIPIF1<0中，SKIPIF1<0，則下列各式一定成立的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<07．（2023·四川成都·石室中學(xué)?？寄M預(yù)測(cè)）設(shè)SKIPIF1<0為等差數(shù)列SKIPIF1<0的前n項(xiàng)和，且SKIPIF1<0，都有SKIPIF1<0，若SKIPIF1<0，則（

）A．SKIPIF1<0的最小值是SKIPIF1<0 B．SKIPIF1<0的最小值是SKIPIF1<0C．SKIPIF1<0的最大值是SKIPIF1<0 D．SKIPIF1<0的最大值是SKIPIF1<08．（2023·陜西咸陽(yáng)·武功縣普集高級(jí)中學(xué)?？寄M預(yù)測(cè)）已知數(shù)列SKIPIF1<0中，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0成等差數(shù)列.若SKIPIF1<0，那么SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<09．（多選題）（2023·安徽安慶·安徽省桐城中學(xué)?？级＃┮阎猄KIPIF1<0為等差數(shù)列，前SKIPIF1<0項(xiàng)和為SKIPIF1<0，SKIPIF1<0，公差d=?2，則（

）A．SKIPIF1<0=SKIPIF1<0B．當(dāng)n=6或7時(shí)，SKIPIF1<0取得最小值C．?dāng)?shù)列SKIPIF1<0的前10項(xiàng)和為50D．當(dāng)n≤2023時(shí)，SKIPIF1<0與數(shù)列SKIPIF1<0（mN）共有671項(xiàng)互為相反數(shù)．10．（多選題）（2023·江蘇鹽城·統(tǒng)考三模）已知數(shù)列SKIPIF1<0對(duì)任意的整數(shù)SKIPIF1<0，都有SKIPIF1<0，則下列說(shuō)法中正確的有（

）A．若SKIPIF1<0，則SKIPIF1<0B．若SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0C．?dāng)?shù)列SKIPIF1<0可以是等差數(shù)列D．?dāng)?shù)列SKIPIF1<0可以是等比數(shù)列11．（多選題）（2023·福建泉州·泉州五中?？寄M預(yù)測(cè)）已知等差數(shù)列SKIPIF1<0的公差為SKIPIF1<0，前SKIPIF1<0項(xiàng)和為SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0成等比數(shù)列，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0是SKIPIF1<0的最大值 D．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0是SKIPIF1<0的最小值12．（多選題）（2023·廣東佛山·?？寄M預(yù)測(cè)）已知數(shù)列SKIPIF1<0，下列結(jié)論正確的有（

）A．若SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0B．若SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0C．若SKIPIF1<0，則數(shù)列SKIPIF1<0是等比數(shù)列D．若SKIPIF1<0為等差數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和，則數(shù)列SKIPIF1<0為等差數(shù)列13．（2023·上海黃浦·上海市大同中學(xué)?？既＃┠纤蔚臄?shù)學(xué)家楊輝“善于把已知形狀、大小的幾何圖形的求面積、體積的連續(xù)量問(wèn)題轉(zhuǎn)化為離散量的垛積問(wèn)題”，在他的專著《詳解九章算法·商功》中，楊輝將堆垛與相應(yīng)立體圖形作類比，推導(dǎo)出了三角垛、方垛、芻童垛等的公式，例如三角垛指的是如圖頂層放1個(gè)，第二層放3個(gè)，第三層放6個(gè)，第四層放10個(gè)SKIPIF1<0第n層放SKIPIF1<0個(gè)物體堆成的堆垛，則SKIPIF1<0______．

14．（2023·廣東佛山·華南師大附中南海實(shí)驗(yàn)高中?？寄M預(yù)測(cè)）設(shè)隨機(jī)變量SKIPIF1<0的分布列如下：SKIPIF1<0123456PSKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0其中SKIPIF1<0，SKIPIF1<0，…，SKIPIF1<0構(gòu)成等差數(shù)列，則SKIPIF1<0___________.15．（2023·甘肅張掖·高臺(tái)縣第一中學(xué)校考模擬預(yù)測(cè)）已知等差數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0，公差d為奇數(shù)，且同時(shí)滿足：①SKIPIF1<0存在最大值；②SKIPIF1<0；③SKIPIF1<0．則數(shù)列SKIPIF1<0的一個(gè)通項(xiàng)公式可以為SKIPIF1<0______．（寫(xiě)出滿足題意的一個(gè)通項(xiàng)公式）16．（2023·上海嘉定·上海市嘉定區(qū)第一中學(xué)?？既＃┮阎猄KIPIF1<0，SKIPIF1<0，將數(shù)列SKIPIF1<0與數(shù)列SKIPIF1<0的公共項(xiàng)從小到大排列得到新數(shù)列SKIPIF1<0，則SKIPIF1<0______.17．（2023·湖南·校聯(lián)考模擬預(yù)測(cè)）記等差數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0，已知SKIPIF1<0，SKIPIF1<0．(1)求SKIPIF1<0的通項(xiàng)公式；(2)設(shè)SKIPIF1<0，數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0，若SKIPIF1<0，求m的值．18．（2023·江蘇·校聯(lián)考模擬預(yù)測(cè)）設(shè)數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0，且滿足SKIPIF1<0.(1)求數(shù)列SKIPIF1<0的通項(xiàng)公式；(2)證明：數(shù)列SKIPIF1<0中的任意不同的三項(xiàng)均不能構(gòu)成等差數(shù)列.19．（2023·浙江·校聯(lián)考模擬預(yù)測(cè)）已知正項(xiàng)等比數(shù)列SKIPIF1<0和數(shù)列SKIPIF1<0，滿足SKIPIF1<0是SKIPIF1<0和SKIPIF1<0的等差中項(xiàng)，SKIPIF1<0.(1)證明：數(shù)列SKIPIF1<0是等差數(shù)列，(2)若數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)積SKIPIF1<0滿足SKIPIF1<0，記SKIPIF1<0，求數(shù)列SKIPIF1<0的前20項(xiàng)和.20．（2023·安徽·校聯(lián)考模擬預(yù)測(cè)）已知數(shù)列SKIPIF1<0滿足：SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，從第二項(xiàng)開(kāi)始，每一項(xiàng)與前一項(xiàng)的差構(gòu)成等差數(shù)列．(1)求SKIPIF1<0；(2)設(shè)SKIPIF1<0，若SKIPIF1<0恒成立，求SKIPIF1<0的取值范圍．1．（2020?新課標(biāo)Ⅱ）北京天壇的圜丘壇為古代祭天的場(chǎng)所，分上、中、下三層．上層中心有一塊圓形石板（稱為天心石），環(huán)繞天心石砌9塊扇面形石板構(gòu)成第一環(huán)，向外每環(huán)依次增加9塊．下一層的第一環(huán)比上一層的最后一環(huán)多9塊，向外每環(huán)依次也增加9塊．已知每層環(huán)數(shù)相同，且下層比中層多729塊，則三層共有扇面形石板（不含天心石）SKIPIF1<0SKIPIF1<0A．3699塊 B．3474塊 C．3402塊 D．3339塊2．（2020?北京）在等差數(shù)列SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0．記SKIPIF1<0，2，SKIPIF1<0，則數(shù)列SKIPIF1<0SKIPIF1<0A．有最大項(xiàng)，有最小項(xiàng) B．有最大項(xiàng)，無(wú)最小項(xiàng) C．無(wú)最大項(xiàng)，有最小項(xiàng) D．無(wú)最大項(xiàng)，無(wú)最小項(xiàng)3．（2022?上海）已知等差數(shù)列SKIPIF1<0的公差不為零，SKIPIF1<0為其前SKIPIF1<0項(xiàng)和，若SKIPIF1<0，則SKIPIF1<0，2，SKIPIF1<0，SKIPIF1<0中不同的數(shù)值有個(gè)．4．（2022?乙卷（文））記SKIPIF1<0為等差數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和．若SKIPIF1<0，則公差SKIPIF1<0．5．（2021?上海）已知等差數(shù)列SKIPIF1<0的首項(xiàng)為3，公差為2，則SKIPIF1<0．6．（2020?上海）已知數(shù)列SKIPIF1<0是公差不為零的等差數(shù)列，且SKIPIF1<0，則SKIPIF1<0．7．（2020?海南）將數(shù)列SKIPIF1<0與SKIPIF1<0的公共項(xiàng)從小到大排列得到數(shù)列SKIPIF1<0，則SKIPIF1<0的前SKIPIF1<0