2024-2025學(xué)年新教材高中數(shù)學(xué)第五章數(shù)列5.2等差數(shù)列5.2.2.2等差數(shù)列習(xí)題課課時(shí)素養(yǎng)檢測(cè)含解析新人教B版選擇性必修第三冊(cè)

PAGE六等差數(shù)列習(xí)題課(30分鐘60分)一、選擇題(每小題5分,共30分,多選題全部選對(duì)得5分,選對(duì)但不全的得3分,有選錯(cuò)的得0分)1.若數(shù)列{an}為等差數(shù)列,Sn為其前n項(xiàng)和,且a2=3a4-6,則S9等于 ()A.54 B.50 C.27【解析】選C.設(shè){an}公差為d,則a4=a2+2d,所以a2=3(a2+2d)-6,2a2+6d-6=0,a2+3d=3,即a5=3,所以S9=QUOTE=9×a5=27.2.已知等差數(shù)列{an}的公差d≠0,Sn是其前n項(xiàng)和,若a1+a3+a5=-15,a2+a4+a6=-21,則QUOTES3的值是 ()A.-5 B.-QUOTE C.-QUOTE D.-QUOTE【解析】選C.由等差數(shù)列性質(zhì)知3a3=-15,3a4=-21,故a3=-5,a4=-7,則a2=-3.則QUOTES3=QUOTE×QUOTE=QUOTE=-QUOTE.3.已知等差數(shù)列{an}的前n項(xiàng)和Sn=n2+n,則過(guò)P(1,a1),Q(2,a2)兩點(diǎn)的直線的斜率是 ()A.1 B.2 C.3 【解析】選B.因?yàn)镾n=n2+n,所以a1=S1=2,a2=S2-S1=6-2=4.所以過(guò)P,Q兩點(diǎn)直線的斜率k=QUOTE=QUOTE=2.4.若數(shù)列{an}的通項(xiàng)an=2n-6,設(shè)bn=|an|,則數(shù)列{bn}的前7項(xiàng)和為 ()A.14 B.24 C.26 【解析】選C.當(dāng)n≤3時(shí),an≤0,bn=|an|=-an=6-2n,即b1=4,b2=2,b3=0.當(dāng)n>3時(shí),an>0,bn=|an|=an=2n-6,即b4=2,b5=4,b6=6,b7=8.所以數(shù)列{bn}的前7項(xiàng)和為4+2+0+2+4+6+8=26.5.已知等差數(shù)列{an}的前n項(xiàng)和為Sn,a5=5,S5=15,則數(shù)列QUOTE的前100項(xiàng)和為 ()A.QUOTE B.QUOTE C.QUOTE D.QUOTE【解析】選A.因?yàn)閍5=5,S5=15,所以QUOTE=15,所以a1=1.所以d=QUOTE=1,所以an=n.所以QUOTE=QUOTE=QUOTE-QUOTE.則數(shù)列QUOTE的前100項(xiàng)的和為:T100=QUOTE+QUOTE+…+QUOTE=1-QUOTE=QUOTE.6.(多選題)等差數(shù)列{an}中,QUOTE是一個(gè)與n無(wú)關(guān)的常數(shù),則該常數(shù)的可能值為()A.1 B.QUOTE C.2 D.3【解析】選AB.本題考查等差數(shù)列.設(shè)等差數(shù)列{an}的公差為d,則QUOTE=QUOTE為常數(shù),則a1=d或d=0,QUOTE=QUOTE或1.二、填空題(每小題5分,共10分)7.在等差數(shù)列{an}中,a2=3,a3+a4=9,則a1a6=______【解析】因?yàn)閍2=3,a3+a4=9,所以a2+a3+a4=12,即3a3=12,故a3=4,a4=5,所以an=n+1,所以a1a6=2×答案:148.已知數(shù)列{an}滿意an=11-2n,則|a1|+|a2|+|a3|+…+|a8|=________.

【解析】原式=(a1+a2+a3+a4+a5)-(a6+a7+a8)=(9+7+5+3+1)-(-1-3-5)=34.答案:34三、解答題(每小題10分,共20分)9.在等差數(shù)列{an}中,a1=-60,a17=-12,求數(shù)列{|an|}的前n項(xiàng)和.【解析】等差數(shù)列{an}的公差d=QUOTE=QUOTE=3,所以an=a1+(n-1)d=-60+(n-1)×3=3n-63.由an<0,得3n-63<0,即n<21.所以數(shù)列{an}的前20項(xiàng)是負(fù)數(shù),第20項(xiàng)以后的項(xiàng)都為非負(fù)數(shù).設(shè)Sn,Sn′分別表示數(shù)列{an}和{|an|}的前n項(xiàng)和,當(dāng)n≤20時(shí),Sn′=|a1|+|a2|+…+|an|=-a1-a2-…-an=-Sn=-QUOTE=-QUOTEn2+QUOTEn;當(dāng)n>20時(shí)Sn′=-S20+(Sn-S20)=Sn-2S20=-60n+QUOTE×3-2×QUOTE=QUOTEn2-QUOTEn+1260.所以數(shù)列{|an|}的前n項(xiàng)和Sn′=QUOTE10.已知Sn為各項(xiàng)均為正數(shù)的數(shù)列{an}的前n項(xiàng)和,a1∈(0,2),QUOTE+3an+2=6Sn.(1)求{an}的通項(xiàng)公式;(2)設(shè)bn=QUOTE,數(shù)列{bn}的前n項(xiàng)和為Tn,若對(duì)隨意n∈N*,t≤4Tn恒成立,求實(shí)數(shù)t的最大值.【解析】(1)①當(dāng)n=1時(shí),QUOTE+3a1+2=6S1=6a1,即QUOTE-3a1+2=0,又因?yàn)閍1∈(0,2),解得a1=1.②對(duì)隨意n∈N*,由QUOTE+3an+2=6Sn知QUOTE+3an+1+2=6Sn+1,兩式相減,得QUOTE-QUOTE+3(an+1-an)=6an+1,即(an+1+an)(an+1-an-3)=0,由an>0得an+1-an-3=0,即an+1-an=3,所以{an}是首項(xiàng)為1,公差為3的等差數(shù)列,所以an=1+3(n-1)=3n-2.(2)由an=3n-2得bn=QUOTE=QUOTE=QUOTE,所以Tn=b1+b2+…+bn=QUOTE=QUOTE=QUOTE.因?yàn)門n+1-Tn=QUOTE-QUOTE=QUOTE>0,所以Tn+1>Tn,即數(shù)列{Tn}是遞增數(shù)列,所以t≤4Tn,QUOTE≤Tn,QUOTE≤T1=QUOTE,t≤1,所以實(shí)數(shù)t的最大值是1.(35分鐘70分)一、選擇題(每小題5分,共20分,多選題全部選對(duì)得5分,選對(duì)但不全的得3分,有選錯(cuò)的得0分)1.設(shè)公差為-2的等差數(shù)列{an},假如a1+a4+a7+…+a97=50,那么a3+a6+a9+…+a99等于()A.-182 B.-78 C.-148 D.-82【解析】選D.a3+a6+a9+…+a99=(a1+2d)+(a4+2d)+(a7+2d)+…+(a97+2d)=(a1+a4+…+a97)+2d×33=50+2×(-2)×33=-82.2.已知等差數(shù)列{an}的前n項(xiàng)和為Sn,S10=0,且Sn≥-5對(duì)一切n∈N*恒成立,則此等差數(shù)列{an}的公差d的取值范圍是 ()A.QUOTE B.QUOTEC.QUOTE D.QUOTE【解析】選B.由S10=10a1+QUOTEd=0,所以a1=-QUOTEd.又Sn=na1+QUOTE=-QUOTEnd+QUOTEd=QUOTE(n2-10n)=QUOTE[(n-5)2-25].由Sn≥-5對(duì)一切n∈N*恒成立,得dn2-10dn+10≥0對(duì)一切n∈N*恒成立,所以d≥0且Δ≤0,所以0≤d≤QUOTE,所以d的取值范圍是QUOTE.3.已知函數(shù)f(x)是(-1,+∞)上的單調(diào)函數(shù),且函數(shù)y=f(x-2)的圖像關(guān)于直線x=1對(duì)稱,若數(shù)列{an}是公差不為0的等差數(shù)列,且f(a50)=f(a51),則數(shù)列{an}的前100項(xiàng)的和為 ()A.-200 B.-100 C.0 【解析】選B.因?yàn)楹瘮?shù)y=f(x-2)的圖像關(guān)于直線x=1對(duì)稱,則函數(shù)f(x)的圖像關(guān)于直線x=-1對(duì)稱,又因?yàn)楹瘮?shù)f(x)是(-1,+∞)上的單調(diào)函數(shù),{an}是公差不為0的等差數(shù)列,f(a50)=f(a51),所以a50+a51=-2,S100=QUOTE=50(a50+a51)=-100.4.(多選題)設(shè){an}(n∈N*)是等差數(shù)列,Sn是其前n項(xiàng)的和,且S5<S6,S6=S7>S8,則下列結(jié)論正確的是 ()A.d<0B.a7=0C.S9>S5D.S6與S7均為Sn的最大值【解析】選ABD.由S5<S6得a1+a2+…+a5<a1+a2+…+a5+a6,即a6>0,又因?yàn)镾6=S7,所以a1+a2+…+a6=a1+a2+…+a6+a7,所以a7=0,故B正確;同理由S7>S8,得a8<0,因?yàn)閐=a7-a6<0,故A正確;而C選項(xiàng)S9>S5,即a6+a7+a8+a9>0,可得2(a7+a8)>0,由結(jié)論a7=0,a8<0,明顯C選項(xiàng)是錯(cuò)誤的.因?yàn)镾5<S6,S6=S7>S8,所以S6與S7均為Sn的最大值,故D正確.二、填空題(每小題5分,共20分)5.若Sn是等差數(shù)列{an}的前n項(xiàng)和,且QUOTE=QUOTE,則QUOTE=________.

【解析】由QUOTE=QUOTE,QUOTE=QUOTE,2a1=5d,而QUOTE=QUOTE=QUOTE=QUOTE.答案:QUOTE6.已知Sn為等差數(shù)列{an}的前n項(xiàng)和,滿意a2+a8=6,S5=-5,則a6=________,Sn的最小值為_(kāi)_______.

【解析】依題意得:QUOTE解得QUOTE所以a6=-5+10=5,Sn=-5n+QUOTE×2=n2-6n,當(dāng)n=3時(shí),Sn的最小值為-9.答案:5-97.已知數(shù)列{an}中a1=1,a2=2,當(dāng)整數(shù)n>1時(shí),Sn+1+Sn-1=2(Sn+S1)都成立,則S15=________.

【解析】因?yàn)閿?shù)列{an}中,當(dāng)整數(shù)n>1時(shí),Sn+1+Sn-1=2(Sn+S1)都成立?Sn+1-Sn=Sn-Sn-1+2?an+1-an=2(n>1).所以當(dāng)n≥2時(shí),{an}是以2為首項(xiàng),2為公差的等差數(shù)列.所以S15=14a2+QUOTE×2+a1=14×2+QUOTE×2+1=211.答案:2118.已知等差數(shù)列{an}的前n項(xiàng)和為Sn,若1≤a1≤3,3≤a1+S3≤6,則QUOTE的取值范圍是________.

【解析】在等差數(shù)列{an}中,a1+a3=2a2,所以S3=a1+a2+a3=3a2,又3≤a1+S3≤6,所以3≤a1+3a2≤6.由1≤a1≤3得QUOTE≤QUOTE≤1.所以1≤QUOTE≤6,即1≤1+QUOTE≤6,所以0≤QUOTE≤QUOTE.即QUOTE的取值范圍是QUOTE.答案:QUOTE三、解答題(每小題10分,共30分)9.已知數(shù)列{an},an∈N*,Sn是其前n項(xiàng)和,Sn=QUOTE(an+2)2.(1)求證:{an}是等差數(shù)列;(2)設(shè)bn=QUOTEan-30,求數(shù)列{bn}的前n項(xiàng)和的最小值.【解析】(1)當(dāng)n=1時(shí),a1=S1=QUOTE(a1+2)2,解得a1=2.當(dāng)n≥2時(shí),an=Sn-Sn-1=QUOTE(an+2)2-QUOTE(an-1+2)2,即8an=(an+2)2-(an-1+2)2,整理得(an-2)2-(an-1+2)2=0,即(an+an-1)(an-an-1-4)=0.因?yàn)閍n∈N*,所以an+an-1>0,所以an-an-1-4=0,即an-an-1=4(n≥2).故數(shù)列{an}是以2為首項(xiàng),4為公差的等差數(shù)列.(2)設(shè)數(shù)列{bn}的前n項(xiàng)和為Tn,因?yàn)閎n=QUOTEan-30,且由(1)知,an=2+(n-1)×4=4n-2(n∈N*),所以bn=QUOTE(4n-2)-30=2n-31.故數(shù)列{bn}是單調(diào)遞增的等差數(shù)列.令2n-31=0,得n=15QUOTE.因?yàn)閚∈N*,所以當(dāng)n≤15時(shí),bn<0;當(dāng)n≥16時(shí),bn>0,即b1<b2<…<b15<0<b16<b17<….故當(dāng)n=15時(shí),Tn取得最小值,最小值為T15=QUOTE×15=-225.10.已知等差數(shù)列{an}(n∈N*)滿意:a3=7,a5+a7=26,{an}的前n項(xiàng)和為Sn.(1)求an及Sn;(2)令bn=QUOTE,求數(shù)列{bn}的前n項(xiàng)和Tn.【解析】(1)設(shè)等差數(shù)列{an}的首項(xiàng)為a1,公差為d,由于a3=7,a5+a7=26,所以a1+2d=7,2a1+10d=26,解得a1=3,d=2.所以an=2n+1,Sn=n(n+2)(n∈N*).(2)因?yàn)閍n=2n+1,所以QUOTE-1=4n(n+1),所以bn=QUOTE=QUOTE.故Tn=b1+b2+…+bn=QUOTE=QUOTE=QUOTE,所以數(shù)列{bn}的前n項(xiàng)和Tn=QUOTE(n∈N*).【補(bǔ)償訓(xùn)練】數(shù)列{an}中,a1=8,a4=2,且滿意an+2-2an+1+an=0(n∈N*).(1)求數(shù)列{an}的通項(xiàng)公式;(2)設(shè)Hn=|a1|+|a2|+…+|an|,求Hn.【解析】(1)因?yàn)閍n+2-2an+1+an=0.所以an+2-an+1=an+1-an=…=a2-a1.所以{an}是等差數(shù)列且a1=8,a4=2,所以d=-2,an=a1+(n-1)d=10-2n.故an=10-2n(n∈N*).(2)因?yàn)閍n=10-2n,令an=0,得n=5.當(dāng)n>5時(shí),an<0;當(dāng)n=5時(shí),an=0;當(dāng)n<5時(shí),an>0.設(shè)Sn=a1+a2+…+an.所以當(dāng)n>5時(shí),Hn=|a1|+|a2|+…+|an|=a1+a2+…+a5-(a6+a7+…+an)=S5-(Sn-S5)=2S5-Sn=n2-9n+40,當(dāng)n≤5時(shí),Hn=|a1|+|a2|+…+|an|=a1+a2+…+an=9n-n2.所以Hn=QUOTE(n∈N*)