高考數(shù)列真題篇1.【2014高考北京理第5題】設(shè)SKIPIF1<0是公比為SKIPIF1<0的等比數(shù)列，則“SKIPIF1<0”是“SKIPIF1<0為遞增數(shù)列”的（）A．充分而不必要條件B．必要而不充分條件C．充分必要條件D．既不充分也不必要條件2.【2015高考北京，理6】設(shè)SKIPIF1<0是等差數(shù)列.下列結(jié)論中正確的是（）A．若SKIPIF1<0，則SKIPIF1<0B．若SKIPIF1<0，則SKIPIF1<0C．若SKIPIF1<0，則SKIPIF1<0D．若SKIPIF1<0，則SKIPIF1<03.【2016高考浙江理數(shù)】如圖，點(diǎn)列{An}，{Bn}分別在某銳角的兩邊上，且SKIPIF1<0，SKIPIF1<0，（SKIPIF1<0）.若SKIPIF1<0（）A．SKIPIF1<0是等差數(shù)列B．SKIPIF1<0是等差數(shù)列C．SKIPIF1<0是等差數(shù)列D．SKIPIF1<0是等差數(shù)列4.【2016年高考四川理數(shù)】某公司為激勵(lì)創(chuàng)新，計(jì)劃逐年加大研發(fā)資金投入.若該公司2015年全年投入研發(fā)資金130萬元，在此基礎(chǔ)上，每年投入的研發(fā)資金比上一年增長12%，則該公司全年投入的研發(fā)資金開始超過200萬元的年份是()（參考數(shù)據(jù)：lg1.12≈0.05，lg1.3≈0.11，lg2≈0.30）（A）2018年（B）2019年（C）2020年（D）2021年5【2015高考福建，理8】若SKIPIF1<0是函數(shù)SKIPIF1<0的兩個(gè)不同的零點(diǎn)，且SKIPIF1<0這三個(gè)數(shù)可適當(dāng)排序后成等差數(shù)列，也可適當(dāng)排序后成等比數(shù)列，則SKIPIF1<0的值等于（）A．6B．7C．8D．96.【2016高考浙江理數(shù)】設(shè)數(shù)列{an}的前n項(xiàng)和為Sn.若S2=4，an+1=2Sn+1，n∈N*，則a1=，S5=.7、【2016高考新課標(biāo)1卷】設(shè)等比數(shù)列SKIPIF1<0QUOTE滿足a1+a3=10,a2+a4=5,則a1a2…an的最大值為．8、【2015江蘇高考，11】數(shù)列SKIPIF1<0滿足SKIPIF1<0，且SKIPIF1<0（SKIPIF1<0），則數(shù)列SKIPIF1<0的前10項(xiàng)和為9、【2015高考新課標(biāo)2，理16】設(shè)SKIPIF1<0是數(shù)列SKIPIF1<0的前n項(xiàng)和，且SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0________．10、【2014，安徽理12】數(shù)列SKIPIF1<0是等差數(shù)列，若SKIPIF1<0構(gòu)成公比為SKIPIF1<0的等比數(shù)列，則SKIPIF1<0________11、【2015高考安徽，理14】已知數(shù)列SKIPIF1<0是遞增的等比數(shù)列，SKIPIF1<0，則數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和等于.11、【2016高考新課標(biāo)2理數(shù)】SKIPIF1<0為等差數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和，且SKIPIF1<0記SKIPIF1<0，其中SKIPIF1<0表示不超過SKIPIF1<0的最大整數(shù)，如SKIPIF1<0．（Ⅰ）求SKIPIF1<0；（Ⅱ）求數(shù)列SKIPIF1<0的前1000項(xiàng)和．12、【2014高考廣東卷.理.19】(本小題滿分14分)設(shè)數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0，滿足SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0.(1)求SKIPIF1<0.SKIPIF1<0.SKIPIF1<0的值；(2)求數(shù)列SKIPIF1<0的通項(xiàng)公式.13、【2016高考山東理數(shù)】已知數(shù)列SKIPIF1<0的前n項(xiàng)和Sn=3n2+8n，SKIPIF1<0是等差數(shù)列，且SKIPIF1<0（Ⅰ）求數(shù)列SKIPIF1<0的通項(xiàng)公式；（Ⅱ）令SKIPIF1<0求數(shù)列SKIPIF1<0的前n項(xiàng)和Tn.14、【2014湖南20】已知數(shù)列SKIPIF1<0滿足SKIPIF1<0,SKIPIF1<0.(1)若SKIPIF1<0為遞增數(shù)列,且SKIPIF1<0成等差數(shù)列,求SKIPIF1<0的值;(2)若SKIPIF1<0,且SKIPIF1<0是遞增數(shù)列,SKIPIF1<0是遞減數(shù)列,求數(shù)列SKIPIF1<0的通項(xiàng)公式.15、【2015高考山東，理18】設(shè)數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0.已知SKIPIF1<0.（I）求SKIPIF1<0的通項(xiàng)公式；（II）若數(shù)列SKIPIF1<0滿足SKIPIF1<0，求SKIPIF1<0的前n項(xiàng)和SKIPIF1<0.16、【2016高考天津理數(shù)】已知SKIPIF1<0是各項(xiàng)均為正數(shù)的等差數(shù)列，公差為SKIPIF1<0，對(duì)任意的SKIPIF1<0是SKIPIF1<0和SKIPIF1<0的等差中項(xiàng).(Ⅰ)設(shè)SKIPIF1<0，求證：SKIPIF1<0是等差數(shù)列；(Ⅱ)設(shè)SKIPIF1<0，求證：SKIPIF1<017、【2014山東.理19】已知等差數(shù)列SKIPIF1<0的公差為2，前SKIPIF1<0項(xiàng)和為SKIPIF1<0，且SKIPIF1<0成等比數(shù)列.

（Ⅰ）求數(shù)列SKIPIF1<0的通項(xiàng)公式；（Ⅱ）令SKIPIF1<0，求數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和SKIPIF1<0.18、【2016高考新課標(biāo)3理數(shù)】已知數(shù)列SKIPIF1<0QUOTE的前n項(xiàng)和SKIPIF1<0QUOTE，QUOTE其中SKIPIF1<0．（=1\*ROMANI）證明SKIPIF1<0QUOTE是等比數(shù)列，并求其通項(xiàng)公式；（=2\*ROMANII）若SKIPIF1<0QUOTE，求SKIPIF1<0．19、【2014新課標(biāo)，理17】已知數(shù)列SKIPIF1<0滿足SKIPIF1<0=1，SKIPIF1<0.（Ⅰ）證明SKIPIF1<0是等比數(shù)列，并求SKIPIF1<0的通項(xiàng)公式；（Ⅱ）證明：SKIPIF1<0.20、【2015高考四川，理16】設(shè)數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和SKIPIF1<0，且SKIPIF1<0成等差數(shù)列.（1）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SKIPIF1<0，求得SKIPIF1<0成立的n的最小值.21、【2015高考新課標(biāo)1，理17】SKIPIF1<0為數(shù)列{SKIPIF1<0}的前SKIPIF1<0項(xiàng)和.已知SKIPIF1<0＞0，SKIPIF1<0=QUOTESKIPIF1<0.（Ⅰ）求{SKIPIF1<0}的通項(xiàng)公式；（Ⅱ）設(shè)SKIPIF1<0QUOTE,求數(shù)列{SKIPIF1<0}的前SKIPIF1<0項(xiàng)和.22、【2014課標(biāo)Ⅰ，理17】已知數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，其中SKIPIF1<0為常數(shù)，（=1\*ROMANI）證明：SKIPIF1<0；（=2\*ROMANII）是否存在SKIPIF1<0，使得SKIPIF1<0為等差數(shù)列？并說明理由.23、【2015高考安徽，理18】設(shè)SKIPIF1<0，SKIPIF1<0是曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線與x軸交點(diǎn)的橫坐標(biāo).（Ⅰ）求數(shù)列SKIPIF1<0的通項(xiàng)公式；（Ⅱ）記SKIPIF1<0，證明SKIPIF1<0.24、已知數(shù)列{SKIPIF1<0}的首項(xiàng)為1，SKIPIF1<0為數(shù)列SKIPIF1<0的前n項(xiàng)和，SKIPIF1<0，其中q>0，SKIPIF1<0.（Ⅰ）若SKIPIF1<0成等差數(shù)列，求SKIPIF1<0的通項(xiàng)公式；（Ⅱ）設(shè)雙曲線SKIPIF1<0的離心率為SKIPIF1<0，且SKIPIF1<0，證明：SKIPIF1<0.25、【2015高考天津，理18】（本小題滿分13分）已知數(shù)列SKIPIF1<0滿足SKIPIF1<0，且SKIPIF1<0成等差數(shù)列.(I)求SKIPIF1<0的值和SKIPIF1<0的通項(xiàng)公式；(II)設(shè)SKIPIF1<0，求數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和.26、已知等差數(shù)列{}的公差，它的前n項(xiàng)和為，若，且成等比數(shù)列，（Ⅰ）求數(shù)列{}的通項(xiàng)公式；（Ⅱ）若數(shù)列{}的前n項(xiàng)和為，求證：．27、【天津市南開中學(xué)2015屆高三第三次月考（理）試題】已知數(shù)列的前n項(xiàng)和（）,數(shù)列.（Ⅰ）求證:數(shù)列是等差數(shù)列，并求數(shù)列的通項(xiàng)公式；(Ⅱ）設(shè)數(shù)列的前n項(xiàng)和為，證明：且時(shí)，；（Ⅲ）設(shè)數(shù)列滿足，（為非零常數(shù)，），問是否存在整數(shù)，使得對(duì)任意，都有？數(shù)列綜合題1．已知等差數(shù)列滿足：，，的前n項(xiàng)和為． （Ⅰ）求及； （Ⅱ）令bn=（），求數(shù)列的前n項(xiàng)和。2.已知遞增的等比數(shù)列滿足是的等差中項(xiàng)。 （Ⅰ）求數(shù)列的通項(xiàng)公式； （Ⅱ）若是數(shù)列的前項(xiàng)和，求 3.等比數(shù)列為遞增數(shù)列，且，數(shù)列(n∈N※)（1）求數(shù)列的前項(xiàng)和；（2），求使成立的最小值．4．已知數(shù)列{}、{}滿足：.(1)求;(2)求數(shù)列{}的通項(xiàng)公式；(3)設(shè)，求實(shí)數(shù)為何值時(shí)恒成立5．在數(shù)列中，為其前項(xiàng)和，滿足．（I）若，求數(shù)列的通項(xiàng)公式；（II）若數(shù)列為公比不為1的等比數(shù)列，且，求．6．已知數(shù)列中，，，（1）求證：數(shù)列為等比數(shù)列。（2）設(shè)數(shù)列的前項(xiàng)和為，若，求正整數(shù)列的最小值。 7．已知數(shù)列的前n項(xiàng)和為，