课时训练7等差数列一、等差数列通项公式的应用1.等差数列{an}中,a2=-5,d=3,则a5为()A.-4B.4C.5D.6答案:B解析:a5=a1+4d=(a1+d)+3d=a2+3d=-5+3×3=4.2.在数列{an}中,a1=2,2an+1=2an+1,则a101的值为()A.49B.50C.51D.52答案:D解析: 2an+1=2an+1,∴an+1=an+12.∴an+1-an=12.∴数列{an}是首项为2,公差为12的等差数列.∴a101=a1+(101-1)d=2+1002=52.3.(2015福建厦门高二期末,2)已知数列{an}满足a1=1,an=an-1+3(n≥2),则a100等于()A.297B.298C.299D.300答案:B解析:由an=an-1+3(n≥2),得an-an-1=3(n≥2),即数列{an}是以3为公差的等差数列.又a1=1,∴a100=1+(100-1)×3=298.4.若等差数列{an}的公差为整数,首项为19,从第6项开始为负值,则公差为()A.-5B.-4C.-3D.-2答案:B解析:设等差数列{an}的公差为d(d∈Z),依题意得a6=a1+5d=19+5d<0,即d<-195,a5=a1+4d=19+4d≥0,即d≥-194,所以-194≤d<-195,又d∈Z,所以d=-4.5.等差数列{an}中,a2=5,a4=a6+6,则a1=.答案:8解析:由a4=a6+6,得2d=a6-a4=-6,∴d=-3.又 a1=a2-d=5-(-3)=8,∴a1=8.二、等差中项的应用6.(2015福建宁德五校联考,1)已知实数m是1和5的等差中项,则m等于()A.√5B.±√5C.3D.±3答案:C解析:因为实数m是1和5的等差中项,所以2m=1+5=6,则m=3.故选C.7.已知m和2n的等差中项是4,2m和n的等差中项是5,则m和n的等差中项是()A.2B.3C.6D.9答案:B解析:依题意可得m+2n=8,2m+n=10,故3m+3n=18⇒m+n=6,故m和n的等差中项是3.8.若lg2,lg(2x-1),lg(2x+3)成等差数列,则x的值为()A.1B.0或32C.32D.log25答案:D解析:由题意得lg2+lg(2x+3)=2lg(2x-1),所以2(2x+3)=(2x-1)2,解得2x=5或2x=-1(舍去),所以x=log25.三、等差数列的判断与证明9.(2015山东威海高二期中,21)数列{an}中,a1=1,a2=2,且an+2-an=1+(-1)n(n∈N*).令bn=a2n,求证{bn}是等差数列,并求{bn}的通项公式.解:当n≥2时,bn-bn-1=a2n-a2n-2=2,∴{bn}是等差数列,且b1=a2=2,∴bn=2n.10.已知1b+c,1c+a,1a+b成等差数列,求证:a2,b2,c2成等差数列.证明: 1b+c,1c+a,1a+b是等差数列,∴1b+c+1a+b=2c+a.∴(a+b)(c+a)+(b+c)(c+a)=2(a+b)(b+c).∴(c+a)(a+c+2b)=2(a+b)(b+c).∴2ac+2ab+2bc+a2+c2=2ab+2ac+2bc+2b2.∴a2+c2=2b2.∴a2,b2,c2成等差数列.(建议用时:30分钟)1.数列{an}的通项公式an=4n-7,则此数列是()A.公差为4的等差数列B.公差为-7的等差数列C.首项为-7的等差数列D.公差为n的等差数列答案:A解析:an+1-an=4(n+1)-4n=4.故选A.2.等差数列1,-1,-3,…,-89的项数是()A.45B.46C.47D.92答案...
