【推薦】專題14數(shù)列中的最值問題2022版高人一籌之高三數(shù)學(xué)一輪復(fù)習(xí)特色專題訓(xùn)練

一、選擇題1．已知等差數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和是SKIPIF1<0,若SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0最大值是A.SKIPIF1<0B.SKIPIF1<0C.SKIPIF1<0D.SKIPIF1<0【答案】C【解析】由等差數(shù)列的前n項(xiàng)和的公式可得：SKIPIF1<0故SKIPIF1<0則SKIPIF1<0,故在數(shù)列SKIPIF1<0中,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,當(dāng)SKIPIF1<0,所以SKIPIF1<0時(shí),SKIPIF1<0達(dá)到最大值.2．若等差數(shù)列的前項(xiàng)和,則的最小值為A．【答案】D3.已知正項(xiàng)等比數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0,且SKIPIF1<0,則SKIPIF1<0的最小值為A.10B.15C.20D.25【答案】C【解析】由題意可得：SKIPIF1<0,由SKIPIF1<0可得SKIPIF1<0,由等比數(shù)列的性質(zhì)可得：SKIPIF1<0成等比數(shù)列,則SKIPIF1<0,綜上可得：SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等號(hào)成立.綜上可得,則SKIPIF1<0的最小值為20.4.已知數(shù)列SKIPIF1<0的通項(xiàng)公式為SKIPIF1<0,記數(shù)列SKIPIF1<0的前項(xiàng)和為,則使SKIPIF1<0成立的的最大值為A．4B．5C．6D．8【答案】C【解析】SKIPIF1<0,SKIPIF1<0SKIPIF1<0,SKIPIF1<0,…,所以使SKIPIF1<0成立的SKIPIF1<0的最大值為SKIPIF1<0,故選C.5．設(shè)數(shù)列SKIPIF1<0為等差數(shù)列,SKIPIF1<0為其前SKIPIF1<0項(xiàng)和,若SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0的最大值為A.3B.4C.SKIPIF1<0D.SKIPIF1<0【答案】B【解析】∵S4≥10,S5≤15,∴a1+a2+a3+a4≥10,a1+a2+a3+a4+a5≤15,∴a5≤5,a3≤3,a1+4d≤5,a1+2d≤3,

兩式相加得：2（a1+3d）≤8,∴a4≤4,故選B.6.等比數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和SKIPIF1<0（SKIPIF1<0為常數(shù)）,若SKIPIF1<0恒成立,則實(shí)數(shù)SKIPIF1<0的最大值是A.3B.4C.5D.6【答案】C7.正項(xiàng)等比數(shù)列｛an｝中,存在兩項(xiàng)am,an（m,nSKIPIF1<0）使得aman=16a12,且a7=a6+2a5,則SKIPIF1<0+SKIPIF1<0的最小值為A.5B.6C.7D.8【答案】B【解析】∵SKIPIF1<0,∴SKIPIF1<0,又SKIPIF1<0,∴SKIPIF1<0,∴SKIPIF1<0,∴SKIPIF1<0,即SKIPIF1<0,SKIPIF1<0SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0,即SKIPIF1<0時(shí)等號(hào)成立,∴SKIPIF1<0的最小值為6,故選B．8.等差數(shù)列SKIPIF1<0的公差為SKIPIF1<0,關(guān)于SKIPIF1<0的不等式SKIPIF1<0的解集為SKIPIF1<0,則使數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和SKIPIF1<0最大的正整數(shù)SKIPIF1<0的值是A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0【答案】B9.已知等差數(shù)列SKIPIF1<0的公差SKIPIF1<0,且SKIPIF1<0,SKIPIF1<0,SKIPIF1<0成等比數(shù)列,若SKIPIF1<0,SKIPIF1<0為數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和,則SKIPIF1<0的最小值為A．4B．3C．SKIPIF1<0D．2【答案】A【解析】由已知有SKIPIF1<0,所以有SKIPIF1<0,數(shù)列SKIPIF1<0通項(xiàng)公式SKIPIF1<0,所以SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0,即SKIPIF1<0時(shí)等號(hào)成立.故選A.10.已知三個(gè)數(shù),,成等比數(shù)列,其倒數(shù)重新排列后為遞增的等比數(shù)列的前三項(xiàng),則能使不等式成立的自然數(shù)的最大值為A．9B．8C．7D．5【答案】C【解析】因?yàn)槿齻€(gè)數(shù)等比數(shù)列,所以,倒數(shù)重新排列后恰好為遞增的等比數(shù)列的前三項(xiàng),為,公比為,數(shù)列是以為首項(xiàng),為公比的等比數(shù)列,則不等式等價(jià)為,整理,得,故選C．11.設(shè)等差數(shù)列SKIPIF1<0滿足:SKIPIF1<0,公差SKIPIF1<0,若當(dāng)且僅當(dāng)SKIPIF1<0時(shí),SKIPIF1<0的前SKIPIF1<0項(xiàng)和SKIPIF1<0取得最大值,則首項(xiàng)SKIPIF1<0的取值范圍是A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D.SKIPIF1<0【答案】A12.設(shè)數(shù)列SKIPIF1<0首項(xiàng)SKIPIF1<0,SKIPIF1<0為SKIPIF1<0的前SKIPIF1<0項(xiàng)和,若SKIPIF1<0,當(dāng)SKIPIF1<0取最大值時(shí),SKIPIF1<0A．4B．2C．6D．3【答案】D【解析】由題意得SKIPIF1<0,所以SKIPIF1<0SKIPIF1<0當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào),故選D.二、填空題13.將10個(gè)數(shù)1,2,3,…,9,10按任意順序排列在一個(gè)圓圈上,設(shè)其中連續(xù)相鄰的3數(shù)之和為SKIPIF1<0,則SKIPIF1<0的最大值不小于__________．【答案】1814.已知SKIPIF1<0是等比數(shù)列,且SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0的最大值為__________．【答案】SKIPIF1<0【解析】SKIPIF1<0SKIPIF1<0SKIPIF1<0,即SKIPIF1<0的最大值為SKIPIF1<0.15．設(shè)等差數(shù)列SKIPIF1<0滿足SKIPIF1<0,SKIPIF1<0,且SKIPIF1<0有最小值,則這個(gè)最小值為__________．【答案】-12【解析】因?yàn)閿?shù)列SKIPIF1<0是等差數(shù)列,且SKIPIF1<0,所以SKIPIF1<0,SKIPIF1<0是一元二次方程SKIPIF1<0的二根,由SKIPIF1<0得SKIPIF1<0,SKIPIF1<0或SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0取得最小值,由SKIPIF1<0解得SKIPIF1<0,SKIPIF1<0時(shí),SKIPIF1<0取得最小值,此時(shí)SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0取得最小值,由SKIPIF1<0解得SKIPIF1<0,SKIPIF1<0時(shí),SKIPIF1<0取得最小值,此時(shí)SKIPIF1<0,故答案為SKIPIF1<0.16．設(shè)等差數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0,且SKIPIF1<0（SKIPIF1<0是常數(shù),SKIPIF1<0）,SKIPIF1<0,又SKIPIF1<0,數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0,若SKIPIF1<0對(duì)SKIPIF1<0恒成立,則正整數(shù)SKIPIF1<0的最大值是__________.【答案】217.數(shù)列{an}是等差數(shù)列,數(shù)列{bn}滿足bn=anan+1an+2（n∈N*）,設(shè)Sn為{bn}的前n項(xiàng)和．若SKIPIF1<0,則當(dāng)Sn取得最大值時(shí)n的值等于_____．【答案】SKIPIF1<0【解析】設(shè)SKIPIF1<0的公差為SKIPIF1<0,由SKIPIF1<0得SKIPIF1<0,SKIPIF1<0,即SKIPIF1<0,所以SKIPIF1<0,從而可知SKIPIF1<0時(shí),SKIPIF1<0,SKIPIF1<0時(shí),SKIPIF1<0,從而SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,故SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,因?yàn)镾KIPIF1<0,SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0,故SKIPIF1<0中SKIPIF1<0最大,故答案為16.18.已知等比數(shù)列SKIPIF1<0的首項(xiàng)為SKIPIF1<0,公比為SKIPIF1<0,前SKIPIF1<0項(xiàng)和為SKIPIF1<0,則SKIPIF1<0的最大值與最小值之和為__________．【答案】SKIPIF1<0【解析】由等比數(shù)列前n項(xiàng)和公式可得SKIPIF1<0,令SKIPIF1<0,當(dāng)SKIPIF1<0為奇數(shù)時(shí),SKIPIF1<0單調(diào)遞減,SKIPIF1<0,當(dāng)SKIPIF1<0為偶數(shù)時(shí),SKIPIF1<0單調(diào)遞增,SKIPIF1<0,則SKIPIF1<0,即SKIPIF1<0,令SKIPIF1<0,函數(shù)單調(diào)遞減,則：SKIPIF1<0,最大值與最小值之和為SKIPIF1<0.19.等差數(shù)列SKIPIF1<0滿足SKIPIF1<0,則SKIPIF1<0的取值范圍是________.【答案】SKIPIF1<0.三、解答題20．已知數(shù)列SKIPIF1<0的各項(xiàng)為正數(shù),其前SKIPIF1<0項(xiàng)和為SKIPIF1<0滿足SKIPIF1<0,設(shè)SKIPIF1<0.（1）求證：數(shù)列SKIPIF1<0是等差數(shù)列,并求SKIPIF1<0的通項(xiàng)公式；（2）設(shè)數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0,求SKIPIF1<0的最大值.（3）設(shè)數(shù)列SKIPIF1<0的通項(xiàng)公式為SKIPIF1<0,問:是否存在正整數(shù)t,使得SKIPIF1<0SKIPIF1<0成等差數(shù)列？若存在,求出t和m的值；若不存在,請(qǐng)說明理由.21．已知數(shù)列SKIPIF1<0是首項(xiàng)等于SKIPIF1<0且公比不為1的等比數(shù)列,SKIPIF1<0是它的前SKIPIF1<0項(xiàng)和