新高考數(shù)學(xué)二輪復(fù)習(xí)培優(yōu)專題訓(xùn)練專題19 等差數(shù)列與等比數(shù)列基本量的問題（原卷版）

專題19等差數(shù)列與等比數(shù)列基本量的問題1、（2023年全國乙卷數(shù)學(xué)（文））已知SKIPIF1<0為等比數(shù)列，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0______.2、（2023年全國甲卷數(shù)學(xué)（文））記SKIPIF1<0為等差數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和．若SKIPIF1<0，則SKIPIF1<0（

）A．25 B．22 C．20 D．153、（2023年全國甲卷數(shù)學(xué)（文））記SKIPIF1<0為等比數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和．若SKIPIF1<0，則SKIPIF1<0的公比為________．4、（2023年全國甲卷數(shù)學(xué)（理））已知正項(xiàng)等比數(shù)列SKIPIF1<0中，SKIPIF1<0為SKIPIF1<0前n項(xiàng)和，SKIPIF1<0，則SKIPIF1<0（

）A．7 B．9 C．15 D．305、（2023年新高考天津卷）已知SKIPIF1<0為等比數(shù)列，SKIPIF1<0為數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和，SKIPIF1<0，則SKIPIF1<0的值為（

）A．3 B．18 C．54 D．1526、【2022年全國乙卷】已知等比數(shù)列an的前3項(xiàng)和為168，a2?A．14 B．12 C．6 D．37、（2023年新課標(biāo)全國Ⅰ卷）記SKIPIF1<0為數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和，設(shè)甲：SKIPIF1<0為等差數(shù)列；乙：SKIPIF1<0為等差數(shù)列，則（

）A．甲是乙的充分條件但不是必要條件B．甲是乙的必要條件但不是充分條件C．甲是乙的充要條件D．甲既不是乙的充分條件也不是乙的必要條件8、（2023年新課標(biāo)全國Ⅰ卷）設(shè)等差數(shù)列SKIPIF1<0的公差為SKIPIF1<0，且SKIPIF1<0．令SKIPIF1<0，記SKIPIF1<0分別為數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和．(1)若SKIPIF1<0，求SKIPIF1<0的通項(xiàng)公式；(2)若SKIPIF1<0為等差數(shù)列，且SKIPIF1<0，求SKIPIF1<0．9、（2023年新課標(biāo)全國Ⅱ卷）記SKIPIF1<0為等比數(shù)列SKIPIF1<0的前n項(xiàng)和，若SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0（

）．A．120 B．85 C．SKIPIF1<0 D．SKIPIF1<010、（2023年全國乙卷數(shù)學(xué)（文））記SKIPIF1<0為等差數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和，已知SKIPIF1<0．(1)求SKIPIF1<0的通項(xiàng)公式；(2)求數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和SKIPIF1<0．11、【2022年全國甲卷】記Sn為數(shù)列an的前n項(xiàng)和．已知(1)證明：an(2)若a4,a題組一、等差、等比數(shù)列的基本量的問題1-1、（2023·江蘇南通·統(tǒng)考模擬預(yù)測）傳說國際象棋發(fā)明于古印度，為了獎賞發(fā)明者，古印度國王讓發(fā)明者自己提出要求，發(fā)明者希望國王讓人在他發(fā)明的國際象棋棋盤上放些麥粒，規(guī)則為：第一個格子放一粒，第二個格子放兩粒，第三個格子放四粒，第四個格子放八粒……依此規(guī)律，放滿棋盤的64個格子所需小麥的總重量大約為（

）噸.（1kg麥子大約20000粒，lg2=0.3）A．105 B．107 C．1012 D．10151-2、（2023·黑龍江·黑龍江實(shí)驗(yàn)中學(xué)校考一模）（多選題）已知SKIPIF1<0是等比數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和，且SKIPIF1<0，則下列說法正確的是（

）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<01-3、（2023·江蘇徐州·徐州市第七中學(xué)?？家荒＃┮阎?xiàng)等差數(shù)列SKIPIF1<0滿足SKIPIF1<0，且SKIPIF1<0是SKIPIF1<0與SKIPIF1<0的等比中項(xiàng)，則SKIPIF1<0的前SKIPIF1<0項(xiàng)和SKIPIF1<0___________.1-4、（2023·云南紅河·統(tǒng)考一模）在數(shù)列SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0為等比數(shù)列，則SKIPIF1<0____________．題組二、等差、等比數(shù)列的判斷與證明2-1、（2023·安徽蚌埠·統(tǒng)考三模）（多選題）已知等差數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0，等比數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)積為SKIPIF1<0，則下列結(jié)論正確的是(

)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是等比數(shù)列 D．?dāng)?shù)列SKIPIF1<0是等差數(shù)列2-2、（2023·重慶·統(tǒng)考三模）（多選題）對于數(shù)列SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0，則下列說法正確的是（

）A．SKIPIF1<0 B．?dāng)?shù)列SKIPIF1<0是等差數(shù)列C．?dāng)?shù)列SKIPIF1<0是等差數(shù)列 D．SKIPIF1<02-3、（2023·江蘇連云港·統(tǒng)考模擬預(yù)測）已知數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0，且SKIPIF1<0．(1)證明：數(shù)列SKIPIF1<0是等差數(shù)列；(2)設(shè)數(shù)列SKIPIF1<0的前n項(xiàng)積為SKIPIF1<0，若SKIPIF1<0，求數(shù)列SKIPIF1<0的通項(xiàng)公式．2-4、（2023·安徽宿州·統(tǒng)考一模）在數(shù)列SKIPIF1<0中，SKIPIF1<0，且SKIPIF1<0.(1)令SKIPIF1<0，證明：數(shù)列SKIPIF1<0為等差數(shù)列，并求數(shù)列SKIPIF1<0的通項(xiàng)公式；(2)記數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0，求SKIPIF1<0.1、（2023·浙江溫州·統(tǒng)考三模）已知數(shù)列SKIPIF1<0各項(xiàng)為正數(shù)，SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0，則（

）A．SKIPIF1<0是等差數(shù)列 B．SKIPIF1<0是等比數(shù)列C．SKIPIF1<0是等差數(shù)列 D．SKIPIF1<0是等比數(shù)列2、（2022·山東日照·高三期末）（多選題）數(shù)列SKIPIF1<0的各項(xiàng)均是正數(shù)，SKIPIF1<0，SKIPIF1<0，函數(shù)SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線過點(diǎn)SKIPIF1<0，則下列正確的是（）A．SKIPIF1<0B．?dāng)?shù)列SKIPIF1<0是等比數(shù)列C．?dāng)?shù)列SKIPIF1<0是等比數(shù)列D．SKIPIF1<03、（2021·河北張家口市·高三期末）（多選題）已知數(shù)列的前項(xiàng)和為，下列說法正確的是（）A．若，則是等差數(shù)列B．若，則是等比數(shù)列C．若是等差數(shù)列，則D．若是等比數(shù)列，且，，則4、（2023·河北唐山·統(tǒng)考三模）設(shè)SKIPIF1<0為等比數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0__________.5、（2023·安徽合肥·校聯(lián)考三模）SKIPIF1<0是公差不為零的等差數(shù)列，前SKIPIF1<0項(xiàng)和為SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0成等比數(shù)列，則SKIPIF1<0________．6、（2022·廣東潮州·高三期末）設(shè)SKIPIF1<0是首項(xiàng)為2的等比數(shù)列，SKIPIF1<0是其前n項(xiàng)和．若SKIPIF1<0