新教材2024-2025學(xué)年高中數(shù)學(xué)第1章數(shù)列1.2等差數(shù)列1.2.3等差數(shù)列的前n項(xiàng)和第2課時等差數(shù)列的前n項(xiàng)和的性質(zhì)分層作業(yè)湘教版選擇性必修第一冊

第2課時等差數(shù)列的前n項(xiàng)和的性質(zhì)A級必備學(xué)問基礎(chǔ)練1.已知等差數(shù)列{an}的前11項(xiàng)和S11=88,則a2+a10=()A.16 B.17 C.18 D.192.已知等差數(shù)列{an}的前n項(xiàng)和為Sn,a1=10,公差d=-72,則Sn取得最大值時n的值為(A.3 B.4 C.5 D.63.等差數(shù)列{an}的前n項(xiàng)和為Sn,S5=12,S10=48,則S15=()A.84 B.108 C.144 D.1564.記等差數(shù)列{an}與{bn}的前n項(xiàng)和分別為Sn與Tn,若SnTn=n+1A.8281 B.8182 C.42415.(多選題)已知等差數(shù)列{an}是遞增數(shù)列,其前n項(xiàng)和為Sn,且滿意a7=3a5,則下列結(jié)論正確的是()A.d>0B.a1<0C.當(dāng)n=5時,Sn最小D.當(dāng)Sn>0時,n的最小值為86.(多選題)已知Sn是等差數(shù)列{an}的前n項(xiàng)和,下列選項(xiàng)可能是{Sn}的圖象的是()7.在等差數(shù)列{an}中,a1>0,a10a11<0,若此數(shù)列的前10項(xiàng)和S10=36,前18項(xiàng)和S18=12,則數(shù)列{|an|}的前18項(xiàng)和T18=.

8.若等差數(shù)列{an}的首項(xiàng)為a1=2022,試寫出一個使該數(shù)列的前n項(xiàng)和有最大值的數(shù)列的通項(xiàng)公式,該通項(xiàng)公式為.

9.設(shè){an}為等差數(shù)列,Sn為數(shù)列{an}的前n項(xiàng)和,已知S7=7,S15=75,Tn為數(shù)列Snn的前n項(xiàng)和,求TB級關(guān)鍵實(shí)力提升練10.已知等差數(shù)列{an}的前n項(xiàng)和為Sn,若S3=S10,S6=Sk,則k的值是()A.6 B.7C.8 D.911.已知數(shù)列{an}為等差數(shù)列,Sn為其前n項(xiàng)和.若a1=2024,且S20202020-S20192019=3,則A.1×20212 B.2×20212C.3×20212 D.4×2021212.已知等差數(shù)列{an}是遞減數(shù)列,且滿意|a1|=|a9|,則數(shù)列{an}的前n項(xiàng)和最大時,n=()A.4或5 B.5或6C.7 D.813.(多選題)設(shè)等差數(shù)列{an}的前n項(xiàng)和為Sn,公差為d,已知a3=12,S12>0,S13<0,則下列結(jié)論正確的有()A.a6+a7<0B.a7<0C.d可以取負(fù)整數(shù)D.對隨意n∈N+,有Sn≤S614.(多選題)已知等差數(shù)列{an}是遞減數(shù)列,且前n項(xiàng)和為Sn,若S7=S11,則()A.a10>0B.當(dāng)n=9時,Sn最大C.S17>0D.S19>015.(2024新高考Ⅰ,7)設(shè)Sn為數(shù)列{an}的前n項(xiàng)和,設(shè)甲:{an}為等差數(shù)列;乙:{Snn}為等差數(shù)列,則(A.甲是乙的充分條件但不是必要條件B.甲是乙的必要條件但不是充分條件C.甲是乙的充要條件D.甲既不是乙的充分條件也不是乙的必要條件16.設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,假如a1=-5,an+1=an+2,n∈N+,那么S1,S2,S3,S4中最小的為.

17.在等差數(shù)列{an}中,奇數(shù)項(xiàng)之和為44,偶數(shù)項(xiàng)之和為33,若此數(shù)列的項(xiàng)數(shù)為奇數(shù),則這個數(shù)列的中間項(xiàng)是第項(xiàng);若此數(shù)列的項(xiàng)數(shù)為偶數(shù),且公差為-12,則此數(shù)列的項(xiàng)數(shù)為.18.設(shè)Sn是等差數(shù)列{an}的前n項(xiàng)和,a3=7,.從①S6=51,②an=an-1-3(n≥2),③S5=a3a5中任選一個,補(bǔ)充在問題中并作答.

(1)求數(shù)列{an}的通項(xiàng)公式;(2)求數(shù)列{an}的前n項(xiàng)和Sn的最值.C級學(xué)科素養(yǎng)創(chuàng)新練19.(多選題)已知數(shù)列{an}的前n項(xiàng)和為Sn=33n-n2,則下列說法正確的是()A.an=34-2nB.S16為Sn的最小值C.|a1|+|a2|+…+|a16|=272D.|a1|+|a2|+…+|a30|=450

第2課時等差數(shù)列的前n項(xiàng)和的性質(zhì)1.A由等差數(shù)列{an}的性質(zhì)可得a1+a11=a2+a10.由于前11項(xiàng)和S11=88=11(a1+a11)2,因此a1+a11=16,則a2.A∵a1=10,d=-72∴Sn=10n+n(n-1)2×(-72)=-∵函數(shù)y=-74x2+474x的圖象的對稱軸為直線x=4714,且圖象開口向下,∴當(dāng)n=3時,Sn取得最大值3.B由等差數(shù)列前n項(xiàng)和的性質(zhì)可知S5,S10-S5,S15-S10成等差數(shù)列.由等差中項(xiàng)性質(zhì)可知2(S10-S5)=S5+(S15-S10),解得S15=108,故選B.4.C由{an},{bn}均為等差數(shù)列,得a10b55.ABD設(shè)等差數(shù)列{an}的公差為d,因?yàn)閧an}是遞增數(shù)列,所以d>0.因?yàn)閍7=3a5,所以a5+2d=3a5,所以d=a5,所以a1=a5-4d=-3d<0,故A,B正確;又因?yàn)閍4=a5-d=d-d=0,所以S3=S4,且為Sn的最小值,故C錯誤;又因?yàn)镾8=8(a1+a8)2=4(a4+a5)=4a5=4d>0,S7=故選ABD.6.ABC因?yàn)镾n是等差數(shù)列{an}的前n項(xiàng)和,所以Sn=an2+bn(a,b為常數(shù),n∈N+),則其對應(yīng)函數(shù)y=ax2+bx,當(dāng)x∈N+時的函數(shù)值,函數(shù)的圖象是過原點(diǎn)的一條曲線.當(dāng)a=0時,該曲線是過原點(diǎn)的直線,如選項(xiàng)C;當(dāng)a≠0時,該曲線是過原點(diǎn)的拋物線,如選項(xiàng)A,B;選項(xiàng)D中的曲線不過原點(diǎn),不符合題意.故選ABC.7.60由a1>0,a10a11<0,知d<0,且a10>0,a11<0,所以T18=a1+a2+…+a10-a11-a12-…-a18=2S10-S18=60.8.an=2023-n(答案不唯一)9.解設(shè)等差數(shù)列{an}的首項(xiàng)為a1,公差為d,則Sn=na1+12n(n-1)d因?yàn)镾7=7,S15=75,所以7a1所以Sn=n2-5n2所以數(shù)列Snn是等差數(shù)列,其首項(xiàng)為-2,公差為所以Tn=-2n+n(n-1)210.B由題意可得{an}的公差d≠0.∵等差數(shù)列的前n項(xiàng)和Sn=d2n2+(a1-d2)n可看作二次函數(shù)y=d2x2+(a1-d2)x當(dāng)x∈N+時的函數(shù)值,且S∴二次函數(shù)的圖象的對稱軸為直線x=3+102又S6=Sk,∴6+k2=1311.D因?yàn)閿?shù)列{an}為等差數(shù)列,Sn為其前n項(xiàng)和,a1=2024,S20202020-S20192019=3,所以數(shù)列{Snn}是以S11=2024為首項(xiàng),3為公差的等差數(shù)列,所以S20212021=S11+2024×3=2024+202412.A∵等差數(shù)列{an}是遞減數(shù)列,且滿意|a1|=|a9|,∴a1+8d=-a1,∴a1=-4d>0.∴an=a1+(n-1)d=(n-5)d.令an≥0,得n≥5.∴數(shù)列{an}的前n項(xiàng)和最大時,n=4或n=5.故選A.13.BD因?yàn)镾12=12a1+12×112·d>0,S13=13a1+13所以2a1+11d>0,a1+6d<0,即a6+a7>0,a7<0,所以a6>0,所以d<0,所以對隨意n∈N+,有Sn≤S6.由a3=12得a1=12-2d,聯(lián)立2a1+11d>0,a1+6d<0,解得-247<d<-3,故d不能取負(fù)整數(shù).故選BD14.BC由S7=S11,得S11-S7=a8+a9+a10+a11=2(a9+a10)=0,則a9+a10=10.又因?yàn)閧an}是遞減數(shù)列,所以a9>0,a10<0,故A錯誤,B正確;S17=17(a1+a17S19=19(a1+a19)2=15.C(充分性)若{an}為等差數(shù)列,設(shè)其首項(xiàng)為a1,公差為d,則Sn=na1+n(n-1)2d,則Snn=a1+n-12d=(必要性)反之,若{Snn}為等差數(shù)列,設(shè)Snn=An+B,A≠0,則Sn=An2+Bn,a1=S1=A+B.當(dāng)n≥2時,an=Sn-Sn-1=A(2n-1)當(dāng)n=1時也符合上式,故an=2An-A+B,故{an}為等差數(shù)列,則甲是乙的必要條件.綜上,甲是乙的充要條件.故選C.16.S3∵數(shù)列{an}的前n項(xiàng)和為Sn,a1=-5,an+1=an+2,n∈N+,∴數(shù)列{an}是首項(xiàng)為-5,公差為2的等差數(shù)列.∴a1=-5,a2=-3,a3=-1,a4=1.∴S1=-5,S2=-8,S3=-9,S4=-8.∴S1,S2,S3,S4中最小的為S3.17.444若此數(shù)列的項(xiàng)數(shù)為奇數(shù),設(shè)項(xiàng)數(shù)為2n-1,則奇數(shù)項(xiàng)之和S1=a1+a3+…+a2n-1=n(a1偶數(shù)項(xiàng)之和S2=a2+a4+a6+…+a2n-2=(n-1)(a2+a2n-2若此數(shù)列的項(xiàng)數(shù)為偶數(shù),設(shè)項(xiàng)數(shù)為2n,則S1-S2=nd,所以-11=-12n,所以n=22,故此數(shù)列的項(xiàng)數(shù)為4418.解若選①:(1)設(shè)等差數(shù)列{an}的公差為d,由題意可得a3=a1+2d=7,S6(2)由(1)可知,an=3n-2,所以數(shù)列{an}是遞增數(shù)列,故Sn的最小值為S1=1,無最大值.若選②:(1)設(shè)等差數(shù)列{an}的公差為d,由題意可得d=an-an-1=-3,因?yàn)閍3=a1+(3-1)×(-3)=7,解得a1=13,所以an=-3n+16.(2)由(1)可得an=-3n+16,令an=-3n+16≥0,an+1=-3n+13≤0,解得133≤若選③:(1)設(shè)等差數(shù)列{an}的公差為d,由題意可得S5=5(a1+a5)2=5a3=a所以d=a5-所以an=a3+(n-3)d=-n+10.(2)由(1)可知an=-n+10,令an=0,解得n=10,故Sn的最大值為S9=S10=10×(9+019.AC數(shù)列{an}的前n項(xiàng)和為Sn=33n-n2.當(dāng)n=1時,a1=32,當(dāng)n≥2時,an=Sn-Sn-1=33n-n2-