第1课时等差数列的前n项和公式同步练习-2025-2026学年高二上学期数学人教A版选择性必修第二册

4.2.2等差数列的前n项和公式第1课时等差数列的前n项和公式一．选择题1.在等差数列{an}中,若a2=1,a4=5,则数列{an}的前5项和S5等于()A.7 B.15 C.20 D.252.已知等差数列{an}的前n项和Sn=n2+5n,则公差d等于()A.1 B.2 C.5 D.103.已知{an}是等差数列,a1=10,前10项和S10=70,则其公差d等于()A.-23 B.-13 C.13 4.已知数列{an}满足2an=an-1+an+1(n>1),Sn是数列{an}的前n项和,a2,a1021是函数f(x)=x2-6x+5的两个零点,则S1022的值为()A.3066 B.12 C.1020 D.30605.已知{an}是公差d=1的等差数列,Sn为数列{an}的前n项和,若S8=4S4,则a10等于()A.172 B.192 C.10 D6.(2023·全国甲高考)记Sn为等差数列{an}的前n项和.若a2+a6=10,a4a8=45,则S5=()A.25 B.22 C.20 D.157.在等差数列{an}中,如果a1+a2+a3=-24,a18+a19+a20=78,那么此数列的前20项和为()A.160 B.180 C.200 D.2208.(多选题)已知等差数列10,7,4,…,则()A.该数列的通项公式为an=-3n+13B.-25是该数列的第13项C.该数列的前5项和最大D.设该数列为{an},则|a1|+|a2|+|a3|+…+|a8|=489.已知等差数列{an}的前n项和为Sn,S4=40,Sn=210,Sn-4=130,则n等于()A.12 B.14 C.16 D.18二．填空题10.已知{an}是等差数列,Sn是它的前n项和.若S4=20,a4=8,则S8=.

11.在等差数列{an}中,若an=2n+3,其前n项和Sn=an2+bn+c(a,b,c为常数),则a-b+c=.

12.已知等差数列{an}的前n项和为Sn,若OB=a1OA+a200OC,且A,B,C三点共线(该直线不过原点O),则S200=.

13.已知等差数列{an},{bn}的前n项和分别为Sn,Tn,且SnTn=7n+45n三．解答题14.已知数列{an}满足a1=50,an+1=an+2n.(1)求{an}的通项公式;(2)已知数列{bn}的前n项和为an,若bm=50,求正整数m的值.15.已知等差数列{an}的前三项依次为a,4,3a,前n项和为Sn,且Sk=110.(1)求a及k的值;(2)设数列{bn}的通项公式为bn=Snn,求证:数列{bn}是等差数列,并求其前n项和T16.已知数列{an}的所有项均为正数,其前n项和为Sn,且Sn=14an2+(1)求证:{an}是等差数列;(2)求数列{an}的通项公式.17.已知公差大于零的等差数列{an}的前n项和为Sn,且满足a3a4=117,a2+a5=22.(1)求数列{an}的通项公式;(2)若{bn}是等差数列,且bn=Snn+4.2.2等差数列的前n项和公式第1课时等差数列的前n项和公式一．选择题1.B设等差数列{an}的公差为d,则有a故S5=5a1+5×42d=152.B∵a1=S1=6,a1+a2=S2=14,∴a2=8,∴d=a2-a1=2.3.A因为S10=10a1+10×92d=70,a1=所以d=-234.A∵数列{an}满足2an=an-1+an+1(n>1),∴数列{an}为等差数列,又a2,a1021是函数f(x)=x2-6x+5的两个零点,即a2,a1021是方程x2-6x+5=0的两个根,∴a2+a1021=6,∴S1022=(a1+a15.B由题意,可知S8=8a1+8×72d=8a1+28,S4=4a1+4×32d=4a1+6.因为S8=4S4,即8a1+28=4(4a1+6),解得a1=12,所以a10=a1+96.C设等差数列{an}的首项为a1,公差为d,由已知得2a1+6d=10,(a1+3d)(a7.B由题意可知a1+a2+a3=3a2=-24,得a2=-8,由a18+a19+a20=3a19=78,得a19=26,于是S20=10(a1+a20)=10(a2+a19)=10×(-8+26)=180.8.AD依题意知,首项a1=10,公差d=-3,所以该数列的通项公式为an=a1+(n-1)d=10-3(n-1)=-3n+13,故A正确;由an=-3n+13=-25,得n=383≠13,故B不正确由an=-3n+13≥0,得n≤4,由an=-3n+13<0,得n≥5,所以该数列的前4项和最大,故C不正确;数列{an}的前n项和Sn=10n+n(n-1)2×|a1|+|a2|+|a3|+…+|a8|=a1+a2+a3+a4-(a5+a6+a7+a8)=2S4-S8=2×-3×42+23×42−-3×89.B因为Sn-Sn-4=an+an-1+an-2+an-3=80,S4=a1+a2+a3+a4=40,所以4(a1+an)=120,a1+an=30.由Sn=n(a1+an二．填空题10.72设等差数列{an}的公差为d,则由S4=4a1+故S8=8×2+8×72×2=7211.-3因为an=2n+3,所以a1=5,Sn=(5+2n+3)n2=n2+4n,与Sn=an2+bn+c比较,得a=1,b=4,12.100因为A,B,C三点共线(该直线不过原点O),所以a1+a200=1,所以S200=200(a113.5由等差数列的性质,知anbn=S2n-1T2n-1=7(2n-1)+45(2n三．解答题14.解:(1)当n≥2时,an=(an-an-1)+(an-1-an-2)+…+(a3-a2)+(a2-a1)+a1=2(n-1)+2(n-2)+…+2×2+2×1+50=2×(n-1)n2又a1=50=12-1+50,故{an}的通项公式为an=n2-n+50.(2)b1=a1=50,当n≥2时,bn=an-an-1=n2-n+50-[(n-1)2-(n-1)+50]=2n-2,即bn=50当m≥2时,令bm=50,得2m-2=50,解得m=26.又b1=50,故正整数m的值为1或26.15.(1)解:设等差数列{an}的公差为d,则a1=a,a2=4,a3=3a,由已知有a+3a=8,得a1=a=2,d=4-2=2,故Sk=ka1+k(k-1)2·d=2k+k由Sk=110,得k2+k-110=0,解得k=10或k=-11(舍去),故a=2,k=10.(2)证明:由(1)得Sn=n(2+2n)则bn=Snn故bn+1-bn=(n+2)-(n+1)=1,且b1=2,即{bn}是首项为2,公差为1的等差数列,故Tn=n(16.(1)证明:当n=1时,a1=S1=14a12+解得a1=3或a1=-1(舍去).当n≥2时,an=Sn-Sn-1=14(an2+2an-3)-14(an-12+2an-1-3),得4an即(an+an-1)(an-an-1-2)=0.因为an+an-1>0,所以an-an-1=2(n≥2).故{an}是以3为首项,2为公差的等差数列.(2)解:由(1)知an=3+2(n-1)=2n+1.17.解:(1)设等差数列{an}的公差为d,则d>0.∵a3+a4=a2+a5=22,且a3a4=117,∴a3,a4是方程x2-22x+117=