（教學(xué)思想典型題專講）高三數(shù)學(xué)一輪復(fù)習(xí) 等差數(shù)列

等差數(shù)列一、選擇題1.(2013年高考重慶卷)在等差數(shù)列{an}中,a2=1,a4=5,則{an}的前5項(xiàng)和S5等于(B)(A)7 (B)15 (C)20 (D)25解析:∵{an}是等差數(shù)列,∴QUOTE?QUOTE∴S5=5a1+QUOTEd=5×(-1)+10×2=15,故選B.2.(2013年高考福建卷)等差數(shù)列{an}中,a1+a5=10,a4=7,則數(shù)列{an}的公差為(B)(A)1 (B)2 (C)3 (D)4解析:∵a1+a5=2a3=10,∴a3=5,又∵a4=7,∴d=2,故選B.3.(2013天津市新華中學(xué)月考)設(shè)Sn是等差數(shù)列{an}的前n項(xiàng)和,S5=3(a2+a8),則QUOTE的值為(D)(A)QUOTE (B)QUOTE (C)QUOTE (D)QUOTE解析:由S5=3(a2+a8)得,QUOTE=3×2a5,即5a3=6a5,所以QUOTE=QUOTE,故選D.4.(2013金華一中月考)已知Sn是等差數(shù)列{an}的前n項(xiàng)和,S5=3(a2+a8),則QUOTE等于(A)(A)QUOTE (B)QUOTE (C)QUOTE (D)QUOTE解析:由等差數(shù)列的性質(zhì)可知,S5=5a3,a2+a8=2a5,因?yàn)镾5=3(a2+a8),所以5a3=3×2a5,QUOTE=QUOTE,故選A.5.(2013廈門市高三上學(xué)期期末質(zhì)量檢查)在等差數(shù)列{an}中,an>0,且a1+a2+…+a10=30,則a5·a6的最大值等于(C)(A)3 (B)6 (C)9 (D)36解析:∵a1+a2+…+a10=30,即QUOTE=30,a1+a10=6,∴a5+a6=6,∴a5·a6≤QUOTE=9,故選C.6.(2013北京海淀)若數(shù)列{an}滿足:a1=19,an+1=an-3(n∈N*),則數(shù)列{an}的前n項(xiàng)和數(shù)值最大時(shí),n的值為(B)(A)6 (B)7 (C)8 (D)9解析:∵an+1-an=-3(n∈N*),∴數(shù)列{an}是以19為首項(xiàng),-3為公差的等差數(shù)列,∴an=19+(n-1)×(-3)=22-3n.設(shè)前k項(xiàng)和最大,則有QUOTE∴QUOTE∴QUOTE≤k≤QUOTE,∵k∈N*,∴k=7.故滿足條件的n的值為7.故選B.二、填空題7.(2013西安八校聯(lián)考)植樹(shù)節(jié)某班20名同學(xué)在一段直線公路一側(cè)植樹(shù),每人植一棵,相鄰兩棵樹(shù)相距10米.開(kāi)始時(shí)需將樹(shù)苗集中放置在某一樹(shù)坑旁邊,使每位同學(xué)從各自樹(shù)坑出發(fā)前來(lái)領(lǐng)取樹(shù)苗往返所走的路程總和最小,這個(gè)最小值為米.

解析:設(shè)樹(shù)苗放在第i個(gè)樹(shù)坑旁邊(如圖),那么各個(gè)樹(shù)坑到第i個(gè)樹(shù)坑距離的和是s=(i-1)×10+(i-2)×10+…+(i-i)×10+[(i+1)-i]×10+…+(20-i)×10=10×[i×i-QUOTE-i×(20-i)+QUOTE]=10(i2-21i+210),所以當(dāng)i=10或11時(shí),s的值最小,最小值是1000,所以往返路程的最小值是2013米.答案:20138.已知數(shù)列{an}中,a1=1且QUOTE=QUOTE+QUOTE(n∈N*),則a10=.

解析:由QUOTE=QUOTE+QUOTE知,數(shù)列QUOTE為等差數(shù)列,則QUOTE=1+QUOTE(n-1),即an=QUOTE.∴a10=QUOTE=QUOTE.答案:QUOTE9.(2013煙臺(tái)高三質(zhì)檢)由正數(shù)組成的等差數(shù)列{an}和{bn}的前n項(xiàng)和分別為Sn和Tn,且QUOTE=QUOTE,則QUOTE=.

解析:由QUOTE=QUOTE=QUOTE,∴取n=3,則有QUOTE=QUOTE=QUOTE.答案:QUOTE三、解答題10.已知等差數(shù)列{an}的前n項(xiàng)和為Sn,且滿足a2+a4=14,S7=70.(1)求數(shù)列{an}的通項(xiàng)公式;(2)設(shè)bn=QUOTE,則數(shù)列{bn}的最小項(xiàng)是第幾項(xiàng)?并求出該項(xiàng)的值.解:(1)設(shè)公差為d,則有QUOTE即QUOTE解得QUOTE所以an=3n-2.(2)數(shù)列{bn}的最小項(xiàng)是第4項(xiàng),Sn=QUOTE[1+(3n-2)]=QUOTE,所以bn=QUOTE=3n+QUOTE-1≥2QUOTE-1=23.當(dāng)且僅當(dāng)3n=QUOTE,即n=4時(shí)取等號(hào),故數(shù)列{bn}的最小項(xiàng)是第4項(xiàng),該項(xiàng)的值為23.11.已知數(shù)列{an}的各項(xiàng)均為正數(shù),前n項(xiàng)和為Sn,且滿足2Sn=QUOTE+n-4(n∈N*).(1)求證:數(shù)列{an}為等差數(shù)列;(2)求數(shù)列{an}的通項(xiàng)公式.(1)證明:當(dāng)n=1時(shí),有2a1=QUOTE+1-4,即QUOTE-2a1-3=0,解得a1=3(a1=-1舍去).當(dāng)n≥2時(shí),有2Sn-1=QUOTE+n-5,又2Sn=QUOTE+n-4,兩式相減得2an=QUOTE-QUOTE+1,即QUOTE-2an+1=QUOTE,也即(an-1)2=QUOTE,因此an-1=an-1或an-1=-an-1.若an-1=-an-1,則an+an-1=1.而a1=3,所以a2=-2,這與數(shù)列{an}的各項(xiàng)均為正數(shù)相矛盾,所以an-1=an-1,即an-an-1=1,因此數(shù)列{an}為等差數(shù)列.(2)解:由(1)知a1=3,d=1,所以數(shù)列{an}的通項(xiàng)公式an=3+(n-1)×1=n+2,即an=n+2.12.(2013南充市第一次適應(yīng)性考試)設(shè)數(shù)列{an}的各項(xiàng)都為正數(shù),其前n項(xiàng)和為Sn,已知對(duì)任意n∈N*,Sn是QUOTE和an的等差中項(xiàng).(1)證明數(shù)列{an}為等差數(shù)列,并求數(shù)列{an}的通項(xiàng)公式;(2)證明:QUOTE+QUOTE+…+QUOTE<2.(1)解:由已知,2Sn=QUOTE+an,且an>0.當(dāng)n=1時(shí),2a1=QUOTE+a1,解得a1=1.當(dāng)n≥2時(shí),有2Sn-1=QUOTE+an-1.于是2Sn-2Sn-1=QUOTE-QUOTE+an-an-1,即2an=QUOTE-QUOTE+an-an-1,于是QUOTE-QUOTE=an+an-1,即(an+an-1)(an-an-1)=an+an-1.因?yàn)閍n+an-1>0,所以an-an-1=1(n≥2).故數(shù)列{an}是首項(xiàng)為1,公差為1的等差數(shù)列,且an=n.(