專題6-1等差數(shù)列，等比數(shù)列中性質(zhì)應(yīng)用（選填）目錄TOC\o"1-1"\h\u專題6-1等差數(shù)列，等比數(shù)列中性質(zhì)應(yīng)用（選填） 1 1題型一：等差（等比）數(shù)列中項 1題型二：等差（等比）數(shù)列下角標(biāo)和性質(zhì) 5題型三：等差（等比）數(shù)列單調(diào)性問題 9等比數(shù)列的單調(diào)性 13題型四：等差（等比）數(shù)列中最大（小）項 16題型五：等差（等比）數(shù)列奇偶項問題 21題型六：等差（等比）數(shù)列片段和性質(zhì) 26題型七：兩個等差數(shù)列前SKIPIF1<0項和之比問題 31 36一、單選題 36二、多選題 42三、填空題 44題型一：等差（等比）數(shù)列中項【典例分析】例題1．（2022·四川·廣安二中模擬預(yù)測（文））已知數(shù)列SKIPIF1<0是等比數(shù)列，且SKIPIF1<0，SKIPIF1<0，SKIPIF1<0成等差數(shù)列，則公比SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．1例題2．（2022·江蘇省響水中學(xué)高二期中）正項等比數(shù)列SKIPIF1<0中，SKIPIF1<0是SKIPIF1<0與SKIPIF1<0的等差中項，若SKIPIF1<0，則SKIPIF1<0（

）A．4 B．8 C．32 D．64例題3．（2022·全國·高三專題練習(xí)）已知等差數(shù)列SKIPIF1<0的前SKIPIF1<0項利為SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0，1成等比數(shù)列，且SKIPIF1<0，則SKIPIF1<0的公差SKIPIF1<0的取值范圍為______．例題4．（2022·全國·高三專題練習(xí)（文））已知正項等差數(shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0，且SKIPIF1<0，若SKIPIF1<0成等比數(shù)列，則等差數(shù)列的通項公式SKIPIF1<0________．【提分秘籍】等差中項由三個數(shù)SKIPIF1<0，SKIPIF1<0,SKIPIF1<0組成的等差數(shù)列可以看成是最簡單的等差數(shù)列.這時,SKIPIF1<0叫做SKIPIF1<0與SKIPIF1<0的等差中項.這三個數(shù)滿足關(guān)系式SKIPIF1<0.等比中項如果SKIPIF1<0，SKIPIF1<0，SKIPIF1<0成等比數(shù)列，那么SKIPIF1<0叫做SKIPIF1<0與SKIPIF1<0的等比中項．即：SKIPIF1<0是SKIPIF1<0與SKIPIF1<0的等比中項?SKIPIF1<0，SKIPIF1<0，SKIPIF1<0成等比數(shù)列?SKIPIF1<0【變式演練】1．（2022·江西·上高二中模擬預(yù)測（理））已知等比數(shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，SKIPIF1<0成等差數(shù)列，則SKIPIF1<0（

）A．SKIPIF1<0或SKIPIF1<0 B．SKIPIF1<0或SKIPIF1<0 C．SKIPIF1<0或SKIPIF1<0 D．SKIPIF1<0或SKIPIF1<02．（2022·安徽省宿州市第二中學(xué)高二期末）已知數(shù)列SKIPIF1<0為等差數(shù)列，且SKIPIF1<0，3，SKIPIF1<0成等比數(shù)列，則SKIPIF1<0為（

）A．1 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03．（2022·吉林·遼源市第五中學(xué)校高二階段練習(xí)）已知SKIPIF1<0，若3是SKIPIF1<0與SKIPIF1<0的等比中項，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．7 C．SKIPIF1<0 D．94．（2022·全國·高三專題練習(xí)）已知在正項等比數(shù)列SKIPIF1<0中SKIPIF1<0成等差數(shù)列，則SKIPIF1<0__________．題型二：等差（等比）數(shù)列下角標(biāo)和性質(zhì)【典例分析】例題1．（2022·河北·衡水市第二中學(xué)高二期中）已知等差數(shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0，若SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0（

）A．2 B．3 C．4 D．6例題2．（2022·吉林·長春市第二中學(xué)高二階段練習(xí)）已知正項等差數(shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0的值為（

）A．3 B．14 C．28 D．42例題3．（2022·浙江·慈溪中學(xué)高二階段練習(xí)）記正項遞增等比數(shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0__________.例題4．（2022·黑龍江·鐵人中學(xué)高二開學(xué)考試）設(shè)函數(shù)SKIPIF1<0，若正項等比數(shù)列SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0______．【提分秘籍】①SKIPIF1<0SKIPIF1<0，則SKIPIF1<0（特別的，當(dāng)SKIPIF1<0，有SKIPIF1<0）②若SKIPIF1<0，則SKIPIF1<0，其中SKIPIF1<0.特別地，若SKIPIF1<0，則SKIPIF1<0，其中SKIPIF1<0.【變式演練】1．（2022·黑龍江·哈師大青岡實驗中學(xué)高三階段練習(xí)）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．150 B．120 C．75 D．602．（2022·黑龍江·大慶實驗中學(xué)模擬預(yù)測（理））正項等比數(shù)列SKIPIF1<0中的項SKIPIF1<0，SKIPIF1<0是函數(shù)SKIPIF1<0的極值點，則SKIPIF1<0（

）A．SKIPIF1<0 B．1 C．SKIPIF1<0 D．23．（2022·遼寧·沈陽市第一二〇中學(xué)高二階段練習(xí)）在等比數(shù)列SKIPIF1<0中，SKIPIF1<0，函數(shù)SKIPIF1<0，則SKIPIF1<0等于（

）A．36 B．34 C．38 D．2124．（2022·全國·高二單元測試）正項遞增等比數(shù)列SKIPIF1<0，前n項的和為SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0＝__．5．（2022·吉林遼源·高二期末）已知數(shù)列SKIPIF1<0是等差數(shù)列，數(shù)列SKIPIF1<0是等比數(shù)列，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0______.題型三：等差（等比）數(shù)列單調(diào)性問題【典例分析】例題1．（2022·北京交通大學(xué)附屬中學(xué)高二期中）已知等差數(shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0，若SKIPIF1<0，且SKIPIF1<0，則下列說法中正確的是（

）A．SKIPIF1<0為遞增數(shù)列 B．當(dāng)且僅當(dāng)SKIPIF1<0時，SKIPIF1<0有最大值C．不等式SKIPIF1<0的解集為SKIPIF1<0 D．不等式SKIPIF1<0的解集為無限集例題2．（2022·浙江·金華市外國語學(xué)校高二開學(xué)考試）已知等差數(shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0，下列結(jié)論正確的是（

）A．?dāng)?shù)列SKIPIF1<0是遞增數(shù)列 B．SKIPIF1<0C．當(dāng)SKIPIF1<0取得最大值時，SKIPIF1<0 D．SKIPIF1<0例題3．（多選）（2022·河南·高三階段練習(xí)（理））各項均為正數(shù)的等比數(shù)列SKIPIF1<0的前n項積為SKIPIF1<0，若SKIPIF1<0，公比SKIPIF1<0，則下列命題錯誤的是（

）A．若SKIPIF1<0，則必有SKIPIF1<0 B．若SKIPIF1<0，則必有SKIPIF1<0是SKIPIF1<0中最大的項C．若SKIPIF1<0，則必有SKIPIF1<0 D．若SKIPIF1<0，則必有SKIPIF1<0例題4．（2022·全國·高三專題練習(xí)）已知等差數(shù)列SKIPIF1<0是遞增數(shù)列，且SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的取值范圍為___________.5．（2022·江蘇南通·高三期中）試寫出一個無窮等比數(shù)列SKIPIF1<0，同時滿足①SKIPIF1<0；②數(shù)列SKIPIF1<0單調(diào)遞減；③數(shù)列SKIPIF1<0不具有單調(diào)性，則當(dāng)SKIPIF1<0時，SKIPIF1<0__________.【提分秘籍】若數(shù)列SKIPIF1<0滿足對一切正整數(shù)SKIPIF1<0，都有SKIPIF1<0（或者SKIPIF1<0），則稱數(shù)列SKIPIF1<0為遞增數(shù)列（遞減數(shù)列）；等差數(shù)列的單調(diào)性①當(dāng)SKIPIF1<0，等差數(shù)列SKIPIF1<0為遞增數(shù)列②當(dāng)SKIPIF1<0，等差數(shù)列SKIPIF1<0為遞減數(shù)列③當(dāng)SKIPIF1<0，等差數(shù)列SKIPIF1<0為常數(shù)列等比數(shù)列的單調(diào)性已知等比數(shù)列SKIPIF1<0的首項為SKIPIF1<0，公比為SKIPIF1<0（1）當(dāng)SKIPIF1<0或SKIPIF1<0時，等比數(shù)列SKIPIF1<0為遞增數(shù)列；（2）當(dāng)SKIPIF1<0或SKIPIF1<0時，等比數(shù)列SKIPIF1<0為遞減數(shù)列；（3）當(dāng)SKIPIF1<0時，等比數(shù)列SKIPIF1<0為常數(shù)列（SKIPIF1<0）（4）當(dāng)SKIPIF1<0時，等比數(shù)列SKIPIF1<0為擺動數(shù)列.【變式演練】1．（2022·陜西·渭南市瑞泉中學(xué)高二階段練習(xí)）設(shè)等比數(shù)列SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的最大值為（

）A．32 B．16 C．128 D．642．（2022·全國·高三專題練習(xí)）已知等比數(shù)列SKIPIF1<0的公比為q，且SKIPIF1<0，則“SKIPIF1<0”是“SKIPIF1<0是遞增數(shù)列”的（

）A．充分不必要條件 B．必要不充分條件 C．充要條件 D．既不充分也不必要條件3．（多選）（2022·江蘇·南京市天印高級中學(xué)高三期中）已知等比數(shù)列SKIPIF1<0的公比為SKIPIF1<0，前SKIPIF1<0項積為SKIPIF1<0，若SKIPIF1<0，且SKIPIF1<0，則下列命題正確的是（

）A．SKIPIF1<0 B．當(dāng)且僅當(dāng)SKIPIF1<0時，SKIPIF1<0取得最大值C．SKIPIF1<0 D．SKIPIF1<04．（2022·全國·高三專題練習(xí)）已知等差數(shù)列{SKIPIF1<0}的前n項和是SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則數(shù)列{|SKIPIF1<0|}中值最小的項為第___項．5．（2022·陜西·西安市雁塔區(qū)第二中學(xué)高一階段練習(xí)）在等差數(shù)列SKIPIF1<0中，SKIPIF1<0，則使SKIPIF1<0成立的最大自然數(shù)n為_______題型四：等差（等比）數(shù)列中最大（小）項【典例分析】例題1．（2022·陜西·虢鎮(zhèn)中學(xué)高二階段練習(xí)）設(shè)SKIPIF1<0，則當(dāng)數(shù)列SKIPIF1<0的前SKIPIF1<0項和取得最小值時，SKIPIF1<0的值為（

）A．4 B．5C．4或5 D．5或6例題2．（2022·全國·高三專題練習(xí)）已知等差數(shù)列SKIPIF1<0的通項公式為SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時，數(shù)列SKIPIF1<0的前SKIPIF1<0項和SKIPIF1<0最大.則當(dāng)SKIPIF1<0時，SKIPIF1<0___________.例題3．（2022·福建省寧德第一中學(xué)高二階段練習(xí)）已知首項為4的數(shù)列SKIPIF1<0滿足SKIPIF1<0．(1)證明：數(shù)列SKIPIF1<0是等差數(shù)列．(2)求數(shù)列SKIPIF1<0的通項公式，并求數(shù)列SKIPIF1<0的最小項．【提分秘籍】①求數(shù)列SKIPIF1<0中最大項方法：當(dāng)SKIPIF1<0時，則SKIPIF1<0是數(shù)列最大項；②求數(shù)列SKIPIF1<0中最小項方法：當(dāng)SKIPIF1<0時，則SKIPIF1<0是數(shù)列最小項；③利用單調(diào)性求解【變式演練】1．（2022·全國·高三專題練習(xí)）已知SKIPIF1<0為等差數(shù)列SKIPIF1<0的前SKIPIF1<0項和，且SKIPIF1<0，SKIPIF1<0，則當(dāng)SKIPIF1<0取最大值時，SKIPIF1<0的值為___________.2．（2022·全國·高二期末）已知數(shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0，且SKIPIF1<0(1)證明：SKIPIF1<0是等比數(shù)列；(2)求數(shù)列SKIPIF1<0的通項公式(3)求數(shù)列SKIPIF1<0的通項公式，并求出SKIPIF1<0為何值時，SKIPIF1<0取得最小值，并說明理由．3．（2022·河南·高三階段練習(xí)（理））已知數(shù)列SKIPIF1<0對任意SKIPIF1<0滿足SKIPIF1<0．（1）求數(shù)列SKIPIF1<0的通項公式；（2）設(shè)數(shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0，求使得SKIPIF1<0成立的正整數(shù)SKIPIF1<0的最小值．題型五：等差（等比）數(shù)列奇偶項問題【典例分析】例題1．（2022·上海·位育中學(xué)高二期末）設(shè)等差數(shù)列的項數(shù)SKIPIF1<0為奇數(shù)，則其奇數(shù)項之和與偶數(shù)項之和的比為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0例題2．（2022·全國·高二）已知數(shù)列SKIPIF1<0的前SKIPIF1<0項和SKIPIF1<0，則數(shù)列SKIPIF1<0的前10項中所有奇數(shù)項之和與所有偶數(shù)項之和的比為（

）A．SKIPIF1<0 B．2 C．SKIPIF1<0 D．SKIPIF1<0例題3．（2022·全國·高三專題練習(xí)）已知正項等比數(shù)列SKIPIF1<0共有SKIPIF1<0項，它的所有項的和是奇數(shù)項的和的SKIPIF1<0倍，則公比SKIPIF1<0______.例題4．（2022·江蘇·高二課時練習(xí)）已知等差數(shù)列SKIPIF1<0中，前SKIPIF1<0（SKIPIF1<0為奇數(shù)）項的和為77，其中偶數(shù)項之和為33，且SKIPIF1<0，求通項公式．【變式演練】1．（2022·全國·高三專題練習(xí)）等比數(shù)列的首項為1，項數(shù)是偶數(shù)，所有得奇數(shù)項之和為85，所有的偶數(shù)項之和為170，則這個等比數(shù)列的項數(shù)為（

）A．4 B．6 C．8 D．102．（2022·上海南匯中學(xué)高二期末）在等差數(shù)列SKIPIF1<0中，已知公差SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0__________.3．（2022·全國·高三專題練習(xí)）若數(shù)列SKIPIF1<0滿足SKIPIF1<0（SKIPIF1<0，SKIPIF1<0是不等于SKIPIF1<0的常數(shù)）對任意SKIPIF1<0恒成立，則稱SKIPIF1<0是周期為SKIPIF1<0，周期公差為SKIPIF1<0的“類周期等差數(shù)列”．已知在數(shù)列SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0．(1)求證：SKIPIF1<0是周期為SKIPIF1<0的“類周期等差數(shù)列”，并求SKIPIF1<0的值；(2)若數(shù)列SKIPIF1<0滿足SKIPIF1<0，求SKIPIF1<0的前SKIPIF1<0項和SKIPIF1<0．4．（2022·江蘇·高二課時練習(xí)）一個等差數(shù)列的前12項和為354，前12項中，偶數(shù)項的和與奇數(shù)項的和之比為32∶27，求公差d．5．（2022·江蘇·高二課時練習(xí)）設(shè)等差數(shù)列SKIPIF1<0的前n項和為SKIPIF1<0．已知SKIPIF1<0，SKIPIF1<0．(1)求SKIPIF1<0；(2)求SKIPIF1<0．題型六：等差（等比）數(shù)列片段和性質(zhì)【典例分析】例題1．（2022·全國·高二課時練習(xí)）等差數(shù)列SKIPIF1<0中其前n項和為SKIPIF1<0,SKIPIF1<0則SKIPIF1<0為．A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0例題2．（2022·全國·高二單元測試）設(shè)等比數(shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0例題3．（多選）（2022·全國·高二課時練習(xí)）關(guān)于等差數(shù)列和等比數(shù)列，下列四個選項中正確的有（

）A．若數(shù)列SKIPIF1<0的前n項和SKIPIF1<0（SKIPIF1<0，SKIPIF1<0，SKIPIF1<0為常數(shù)），則數(shù)列SKIPIF1<0為等差數(shù)列B．若數(shù)列SKIPIF1<0的前n項和SKIPIF1<0，則數(shù)列SKIPIF1<0為等比數(shù)列C．?dāng)?shù)列SKIPIF1<0是等差數(shù)列，SKIPIF1<0為前SKIPIF1<0項和，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，…仍為等差數(shù)列D．?dāng)?shù)列SKIPIF1<0是等比數(shù)列，SKIPIF1<0為前SKIPIF1<0項和，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，…仍為等比數(shù)列例題4．（2022·全國·高二課時練習(xí)）記SKIPIF1<0為正項等比數(shù)列SKIPIF1<0的前SKIPIF1<0項和，若SKIPIF1<0，則SKIPIF1<0的最小值為__.【提分秘籍】當(dāng)SKIPIF1<0是等差數(shù)列，記SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0…也成等差數(shù)列，公差為SKIPIF1<0.當(dāng)SKIPIF1<0是等比數(shù)列，記SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0…也成等比數(shù)列，公差為SKIPIF1<0.【變式演練】1．（2022·黑龍江實驗中學(xué)高二階段練習(xí)）公比SKIPIF1<0的等比數(shù)列的前3項，前6項，前9項的和分別為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則下面等式成立的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<02．（2022·全國·高三專題練習(xí)）已知正項等比數(shù)列{an}的前n項和為Sn，且S8-2S4=5，則a9+a10+a11+a12的最小值為（

）A．10 B．15 C．20 D．253．（2022·全國·高二課時練習(xí)）設(shè)等比數(shù)列SKIPIF1<0的前n項和為SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．4 D．54．（2022·上海·高二課時練習(xí)）等差數(shù)列前10項的和為10，第11項至第20項的和為SKIPIF1<0，則第21項至第30項的和是_______．5．（2022·全國·高二課時練習(xí)）已知一個等差數(shù)列SKIPIF1<0的前4項和為32，前8項和為56．(1)求SKIPIF1<0、SKIPIF1<0的值；(2)通過計算觀察，尋找SKIPIF1<0、SKIPIF1<0、SKIPIF1<0、SKIPIF1<0之間的關(guān)系，你發(fā)現(xiàn)什么結(jié)論？(3)根據(jù)上述結(jié)論，請你歸納出對于等差數(shù)列而言的一般結(jié)論，并證明．題型七：兩個等差數(shù)列前SKIPIF1<0項和之比問題【典例分析】例題1．（2022·全國·高三專題練習(xí)）已知兩個等差數(shù)列SKIPIF1<0和SKIPIF1<0的前SKIPIF1<0項和分別為SKIPIF1<0和SKIPIF1<0，且SKIPIF1<0=SKIPIF1<0，則使得SKIPIF1<0為整數(shù)的正整數(shù)SKIPIF1<0的個數(shù)為（

）A．4 B．5 C．6 D．7例題2．（2022·全國·高二課時練習(xí)）兩等差數(shù)列SKIPIF1<0和SKIPIF1<0，前SKIPIF1<0項和分別為SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0例題3．（2022·江蘇·北大附屬宿遷實驗學(xué)校高二期中）已知兩個等差數(shù)列SKIPIF1<0和SKIPIF1<0的前SKIPIF1<0項和分別為SKIPIF1<0和SKIPIF1<0，且SKIPIF1<0=SKIPIF1<0，則SKIPIF1<0的值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0例題4．（2022·全國·高三專題練習(xí)）已知SKIPIF1<0，SKIPIF1<0分別為等差數(shù)列SKIPIF1<0，SKIPIF1<0的前SKIPIF1<0項和，SKIPIF1<0，設(shè)點SKIPIF1<0是直線SKIPIF1<0外一點，點SKIPIF1<0是直線SKIPIF1<0上一點，且SKIPIF1<0SKIPIF1<0，則實數(shù)SKIPIF1<0的值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【提分秘籍】若數(shù)列SKIPIF1<0,SKIPIF1<0均為等差數(shù)列且其前SKIPIF1<0項和分別為SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0【變式演練】1．（2022·全國·高二課時練習(xí)）設(shè)數(shù)列SKIPIF1<0，SKIPIF1<0都是正項等比數(shù)列，SKIPIF1<0，SKIPIF1<0分別為數(shù)列SKIPIF1<0與SKIPIF1<0的前n項和，且SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．（2022·全國·高三專題練習(xí)（理））已知等差數(shù)列SKIPIF1<0的前SKIPIF1<0項和分別為SKIPIF1<0，若對于任意的自然數(shù)SKIPIF1<0，都有SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03．（2022·江蘇·宿遷中學(xué)高二期中）若兩個等差數(shù)列SKIPIF1<0的前n項和分別為An、Bn，且滿足SKIPIF1<0，則SKIPIF1<0的值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<04．（2022·全國·高三專題練習(xí)）已知等差數(shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0，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<05．（2022·廣東·南海中學(xué)高二階段練習(xí)）已知等差數(shù)列SKIPIF1<0、SKIPIF1<0，其前SKIPIF1<0項和分別為SKIPIF1<0、SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<05．（2022·全國·高三專題練習(xí)）已知等差數(shù)列SKIPIF1<0，SKIPIF1<0的前SKIPIF1<0項和分別為SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0______．一、單選題1．（2022·甘肅·高臺縣第一中學(xué)高三階段練習(xí)（文））已知在等比數(shù)列SKIPIF1<0中，SKIPIF1<0，等差數(shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0（

）A．96 B．102 C．118 D．1262．（2022·山東濟寧·高三期中）設(shè)等差數(shù)列SKIPIF1<0的前SKIPIF1<0項的和為SKIPIF1<0，則下列結(jié)論不正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．?dāng)?shù)列SKIPIF1<0的前SKIPIF1<0和為SKIPIF1<03．（2022·廣西玉林·高三階段練習(xí)（理））設(shè)等比數(shù)列SKIPIF1<0的公比為q，其前n項和為SKIPIF1<0，并且滿足條件SKIPIF1<0，則下列結(jié)論正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0的最大值為SKIPIF1<04．（2022·貴州·貴陽六中一模（理））已知數(shù)列SKIPIF1<0的前SKIPIF1<0項和組成的數(shù)列SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則數(shù)列SKIPIF1<0的通項公式為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<05．（2022·福建·莆田第六中學(xué)高二階段練習(xí)）已知等差數(shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0且SKIPIF1<0，則SKIPIF1<0的通項公式為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<06．（2022·四川外國語大學(xué)附屬外國語學(xué)校高三期中）1883年，德國數(shù)學(xué)家康托提出了三分康托集，亦稱康托爾集.下圖是其構(gòu)造過程的圖示，其詳細構(gòu)造過程可用文字描述為：第一步，把閉區(qū)間SKIPIF1<0平均分成三段，去掉中間的一段，剩下兩個閉區(qū)間SKIPIF1<0和SKIPIF1<0；第二步，將剩下的兩個閉區(qū)間分別平均分為三段，各自去掉中間的一段，剩下四段閉區(qū)間：SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0；如此不斷的構(gòu)造下去，最后剩下的各個區(qū)間段就構(gòu)成了三分康托集.若經(jīng)歷SKIPIF1<0步構(gòu)造后，所有去掉的區(qū)間長度和為（

）（注：SKIPIF1<0或SKIPIF1<0或SKIPIF1<0或SKIPIF1<0的區(qū)間長度均為SKIPIF1<0）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<07．（2022·貴州遵義·高三階段練習(xí)（理））數(shù)列SKIPIF1<0滿足：SKIPIF1<0，SKIPIF1<0，記數(shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0，若SKIPIF1<0恒成立，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0