PAGEPAGE1課時(shí)規(guī)范練29等差數(shù)列及其前n項(xiàng)和基礎(chǔ)鞏固組1.由a1=1,d=3確定的等差數(shù)列{an},當(dāng)an=298時(shí),序號(hào)n等于()A.99 B.100 C.96 D.1012.(2024湖南長(zhǎng)郡中學(xué)仿真,6)已知等差數(shù)列{an}滿意an+1+an=4n,則a1=()A.-1 B.1 C.2 D.33.(2024河南商丘二模,3)已知等差數(shù)列{an}的公差為d,且a8+a9+a10=24,則a1·d的最大值為()A. B. C.2 D.44.在等差數(shù)列{an}中,a3+a6=11,a5+a8=39,則公差d為 ()A.-14 B.-7 5.已知等差數(shù)列{an}的前n項(xiàng)和為Sn,且3a3=a6+4,若S5<10,則a2的取值范圍是(A.(-∞,2) B.(-∞,0)C.(1,+∞) D.(0,2)6.已知Sn是等差數(shù)列{an}的前n項(xiàng)和,則2(a1+a3+a5)+3(a8+a10)=36,則S11=()A.66 B.55 C.44 D.337.(2024湖南衡陽(yáng)一模,15)已知數(shù)列{an}前n項(xiàng)和為Sn,若Sn=2an-2n,則Sn=.

8.設(shè)數(shù)列{an}{bn}都是等差數(shù)列,若a1+b1=7,a3+b3=21,則a5+b5=.

9.若數(shù)列{an}的前n項(xiàng)和為Sn,且滿意an+2SnSn-1=0(n≥2),a1=.(1)求證:1S(2)求數(shù)列{an}的通項(xiàng)公式.10.設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,且Sn=2n-1.數(shù)列{bn}滿意b1=2,bn+1-2bn=8an.(1)求數(shù)列{an}的通項(xiàng)公式;(2)證明:數(shù)列bn2n為等差數(shù)列,并求{b綜合提升組11.(2024河北衡水中學(xué)考前押題二,10)已知數(shù)列a1=1,a2=2,且an+2-an=2-2(-1)n,n∈N+,則S2017的值為()A.2016×1010-1 B.1009×2017C.2017×1010-1 D.1009×201612.若數(shù)列{an}滿意: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()A.6 B.7 C.8 D.913.已知等差數(shù)列{an}的前n項(xiàng)和為Sn,若Sm-1=-4,Sm=0,Sm+2=14(m≥2,且m∈N+),則m的值為.

14.已知公差大于零的等差數(shù)列{an}的前n項(xiàng)和為Sn,且滿意a3·a4=117,a2+a5=22.(1)求通項(xiàng)公式an;(2)求Sn的最小值;(3)若數(shù)列{bn}是等差數(shù)列,且bn=Snn創(chuàng)新應(yīng)用組15.(2024湖南長(zhǎng)郡中學(xué)仿真,15)若數(shù)列{an}是正項(xiàng)數(shù)列,且a1+a2+…+an=n2+3n,則a1216.等差數(shù)列{an}的前n項(xiàng)和為Sn,已知S10=0,S15=25,則nSn的最小值為多少?參考答案課時(shí)規(guī)范練29等差數(shù)列及其前n項(xiàng)和1.B依據(jù)等差數(shù)列通項(xiàng)公式an=a1+(n-1)d,有298=1+(n-1)×3,解得n=100,故選B.2.B由題意,當(dāng)n分別取1,2時(shí),a1+a2=4,a3+a2=8,解得公差d=2,故a1=1.故選B.3.C∵a8+a9+a10=24,∴a9=8,即a1+8d=8,∴a1=8-8d,a1·d=(8-8d)d=-8d-122+2≤2,當(dāng)d=12時(shí),a1·d的最大值為2,故選C.4.C∵a3+a6=11,a5+a8=39,則4d=28,解得d=7.故選C.5.A設(shè)公差為d,由3a3=a6+4得3(a2+d)=a2+4d+4,即d=2a2由S5<10得,5(a1+a5)2=5(a2+a4)6.D由等差數(shù)列的性質(zhì)可得2(a1+a3+a5)+3(a8+a10)=6a3+6a9=36,即a1+a11=6.則S11=11(a1+a11)7.n·2n∵Sn=2an-2n=2(Sn-Sn-1)-2n,整理得Sn-2Sn-1=2n,等式兩邊同時(shí)除以2n,則Sn2n-S又S1=2a1-2=a1,可得a1=S1=∴數(shù)列Sn公差為1的等差數(shù)列,∴Sn2n=n,∴Sn=n·8.35∵數(shù)列{an},{bn}都是等差數(shù)列,設(shè)數(shù)列{an}的公差為d1,數(shù)列{bn}的公差為d2,∴a3+b3=a1+b1+2(d1+d2)=21,而a1+b1=7,可得2(d1+d2)=21-7=14.∴a5+b5=a3+b3+2(d1+d2)=21+14=35.9.(1)證明當(dāng)n≥2時(shí),由an+2SnSn-1=0,得Sn-Sn-1=-2SnSn-1,∴1Sn-1S又1S1=1a1=(2)解由(1)可得1Sn=2n,∴Sn=當(dāng)n≥2時(shí),an=Sn-Sn-1=12n-12(n當(dāng)n=1時(shí),a1=12不適合上式故an=110.(1)解當(dāng)n=1時(shí),a1=S1=21-1=1;當(dāng)n≥2時(shí),an=Sn-Sn-1=(2n-1)-(2n-1-1)=2n-1.∵a1=1適合通項(xiàng)公式an=2n-1,∴an=2n-1.(2)證明∵bn+1-2bn=8an,∴bn+1-2bn=2n+2,即bn+12n+1又b12∴bn2∴bn2n=1+2(n-1)=2∴bn=(2n-1)×2n.11.C由題意,當(dāng)n為奇數(shù)時(shí),an+2-an=4,數(shù)列{a2n-1}是首項(xiàng)為1,公差為4的等差數(shù)列,當(dāng)n為偶數(shù)時(shí),an+2-an=0,數(shù)列{a2n-1}是首項(xiàng)為2,公差為0的等差數(shù)列,S2017=(a1+a3+…+a2017)+(a2+a4+…+a2016)=1009+×1009×1008×4+1008×2=2017×1010-1,故選C.12.B∵a1=19,an+1-an=-3,∴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)數(shù)列{an}的前k項(xiàng)和數(shù)值最大,則有ak≥0,ak∴22∴193≤k≤22∵k∈N+,∴k=7.∴滿意條件的n的值為7.13.5∵Sm-1=-4,Sm=0,Sm+2=14,∴am=Sm-Sm-1=4,am+1+am+2=Sm+2-Sm=14.設(shè)數(shù)列{an}的公差為d,則2am+3d=14,∴d=2.∵Sm=a1+am2×m=0,∴a1∴am=a1+(m-1)d=-4+2(m-1)=4,∴m=5.14.解(1)∵數(shù)列{an}為等差數(shù)列,∴a3+a4=a2+a5=22.又a3·a4=117,∴a3,a4是方程x2-22x+117=0的兩實(shí)根.又公差d>0,∴a3<a4,∴a3=9,a4=13,∴a1+2d=9,a1+3d=13,(2)由(1)知a1=1,d=4,∴Sn=na1+n(n-1)2d=2n2∴當(dāng)n=1時(shí),Sn最小,最小值為S1=a1=1.(3)由(2)知Sn=2n2-n,∴bn=Snn+c=2n2-nn+c,∴b1=1∵數(shù)列{bn}是等差數(shù)列,∴2b2=b1+b3,即62+c×2=11+c+153+c∴c=-12(c=0舍去),故c=-115.2n2+6n由a1+a2+…+an=n2+3n,則a1+a2+…+an-1=(n-1)2+3(n-1),兩式相減,可得an=2n+2,當(dāng)n=1時(shí)也成立,則an=(2n+2)2,有ann+1=(2n+2)2n+1=4n+4,ann16.解設(shè)數(shù)列{an}的首項(xiàng)為a1,公差為d,則S10=10a1+10×92d=10a1+45d=S15=15a1+15×142d=15a1+105d=25聯(lián)立①②,得a1=-3,d=