2024-2025學(xué)年高二數(shù)學(xué)選擇性必修第一冊(cè)（配湘教版）1.2.1 第1課時(shí) 等差數(shù)列的概念及通項(xiàng)公式

1.2等差數(shù)列1.2.1等差數(shù)列及其通項(xiàng)公式第1課時(shí)等差數(shù)列的概念及通項(xiàng)公式A級(jí)必備知識(shí)基礎(chǔ)練1.在等差數(shù)列{an}中,若a1=2,a2=4,則a4=()A.6 B.8 C.16 D.322.[2024甘肅臨夏高二期中]已知數(shù)列{an}是等差數(shù)列,a3=2,公差d=-12,則其首項(xiàng)a1=(A.3 B.4C.5 D.63.在等差數(shù)列{an}中,a3+a9=32,a2=4,則a10=()A.25 B.28 C.31 D.344.若等差數(shù)列{an}的公差d=2,a8∶a7=7∶8,則a1=()A.-15 B.-28 C.15 D.285.已知等差數(shù)列{an}的各項(xiàng)都不相等,a1=2,且a4+a8=a32,則公差d=(A.1 B.12 C.2 D.2或6.(多選題)已知數(shù)列{an}為等差數(shù)列,則下列說法正確的是()A.數(shù)列{an+b}(b為常數(shù))是等差數(shù)列B.數(shù)列{-an}是等差數(shù)列C.數(shù)列1an是等差數(shù)列D.an+1是an與an+2的等差中項(xiàng)7.已知等差數(shù)列{an}的前3項(xiàng)依次是-1,a-1,1,則a=,通項(xiàng)公式an=.

B級(jí)關(guān)鍵能力提升練8.若a≠b,則等差數(shù)列a,x1,x2,b的公差是()A.b-a B.b-a2 C.b9.在數(shù)列{an}中,a1=-1,a3=3,an+2=2an+1-an(n∈N+),則a10=()A.10 B.17 C.21 D.3510.[2024甘肅白銀高二期中](多選題)若各項(xiàng)均為正數(shù)的數(shù)列{an}是等差數(shù)列,且a2=5,則()A.當(dāng)a3=7時(shí),a7=15B.a4的取值范圍是[5,15)C.當(dāng)a7為整數(shù)時(shí),a7的最大值為29D.公差d的取值范圍是(0,5)11.在數(shù)列{xn}中,2xn=1xn-1+1xn+1(n≥2),且x2=212.[2024甘肅隴南高二期末]已知數(shù)列{log2(an-1)}(n∈N+)為等差數(shù)列,且a1=3,a3=9,則數(shù)列{an}的通項(xiàng)公式為.

13.已知數(shù)列{an}滿足an=2an-1+2n-1(n≥2),且a4=81.(1)求數(shù)列{an}的前三項(xiàng)a1,a2,a3.(2)是否存在一個(gè)實(shí)數(shù)λ,使得數(shù)列an+λ2n為等差數(shù)列?若存在,求出λ的值;若不存在(3)求數(shù)列{an}的通項(xiàng)公式.C級(jí)學(xué)科素養(yǎng)創(chuàng)新練14.在等差數(shù)列{an}中,a2,a8是方程x2+mx-8=0的兩根,若a4+a6=a52+1,則實(shí)數(shù)m的值為(A.-6 B.-2C.2 D.6

第1課時(shí)等差數(shù)列的概念及通項(xiàng)公式1.B因?yàn)樵诘炔顢?shù)列{an}中,a1=2,a2=4,所以公差d=a2-a1=4-2=2,則a4=a1+3d=2+3×2=8,故選B.2.A∵a3=a1+2d=2,公差d=-12則a1=a3-2d=2+1=3.故選A.3.B因?yàn)樵诘炔顢?shù)列{an}中,a3+a9=32,a2=4,所以2a1+10d=32,a1+d=4,解得a1=1,d=3,所以a10=a1+9d=28,故選B.4.B設(shè)a8=7k,a7=8k,k∈R,則a8-a7=7k-8k=-k=2,則k=-2.即a7=-16,故a1=a7-6d=-16-12=-28,故選B.5.B∵a1=2,且a4+a8=a32,∴2+3d+2+7d=(2+2d)整理可得4d2-2d=0.∵d≠0,∴d=12.故選B6.ABD記數(shù)列{an}的公差為d,則an+1-an=d.(an+1+b)-(an+b)=d,故A正確;(-an+1)-(-an)=-(an+1-an)=-d,所以數(shù)列{-an}是等差數(shù)列,故B正確;1an+1-1an=an-an根據(jù)等差數(shù)列的定義可知2an+1=an+an+2,所以an+1是an與an+2的等差中項(xiàng),故D正確.故選ABD.7.1n-2因?yàn)?1,a-1,1構(gòu)成等差數(shù)列,所以2(a-1)=-1+1=0,解得a=1.因?yàn)閍1=-1,d=1,所以an=n-2.8.C由等差數(shù)列的通項(xiàng)公式,得b=a+(4-1)d,所以d=b-9.B∵an+2=2an+1-an(n∈N+),∴an+2+an=2an+1,即數(shù)列{an}是等差數(shù)列.設(shè){an}的公差為d,∵a1=-1,a3=3,∴a3=a1+2d,即3=-1+2d,得d=2.則an=-1+(n-1)×2=2n-3,故a10=2×10-3=17.10.ABC當(dāng)a2=5,a3=7時(shí),公差d=2,a7=a3+4d=7+8=15,故A正確;因?yàn)閧an}各項(xiàng)均為正數(shù),所以a1=5-d>0,即d<5,且d≥0,所以公差d的取值范圍是[0,5),故D錯(cuò)誤;因?yàn)閍4=5+2d,所以a4的取值范圍是[5,15),故B正確;a7=5+5d∈[5,30),當(dāng)a7為整數(shù)時(shí),a7的最大值為29,故C正確.故選ABC.11.211在數(shù)列{xn}中,因?yàn)?xn=1xn-1+1xn+1(n≥2),所以數(shù)列1xn是等差數(shù)列.又因?yàn)閤2=23,x4=25,所以數(shù)列1xn的公差d=121x4-1x2=12×12.an=2n+1設(shè)等差數(shù)列{log2(an-1)}的公差為d,由a1=3,a3=9,得log2(a3-1)=log2(a1-1)+2d,解得d=1,所以log2(an-1)=1+(n-1)×1=n,即an=2n+1.13.解(1)由題意知a4=2a3+24-1=81,解得a3=33.同理可得a2=13,a1=5.(2)假設(shè)存在實(shí)數(shù)λ滿足題意,則an+λ2n-an∵an+λ2n-an-1+λ2n-1=an-2an-1(3)由(2)知數(shù)列an-12n是等差數(shù)列,其首項(xiàng)為2,公差為1,則an-12n=2+(n-1)·1,故an14.B因?yàn)閍2,a8是方