设等差数列an的公差d不为零a一等于九d若ak是a一与a二k的等比中项则k等于
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由于为等比数列,只要连续3项就可确定数列的首项和公比!故只需要讨论4项删去某一项后剩3项即可!故只要讨论a1,a2,a3,a4即可!(1)删掉首项:a2,a3,a4a3^2=(a3-d)(a3+d)d
设数列{an}的公差为d,则各项分别为:a1,a1+d,a1+2d,…,a1+(n-1)d,且a1≠0,d≠0,假设去掉第一项,则有(a1+d)(a1+3d)=(a1+2d)2,解得d=0,不合题意;
1.因为等差数列AN的公差d不等于0,a1=2,s9=36,所以36=9*2+1/2*9*8d所以d=1/2所以a3=3,a9=6,由a3,a9,am成等比数列则a9的平方=a3*am,的am=12又
(1)根据题意,设公差为d则a3=a1+2d=2d+1a9=a1+8d=8d+1有(2d+1)^2=8d+1d=1故通项:an=n(2)根据题意,设公比为q则b2=qb3=q^2有q-0.5q^2=0
设等差数列的公差为d,则a3=a5-2d=6-2d,an1=a5+(n1-5)d=6+(n1-5)d.∵a3,a5,an1成等比数列,∴a52=a3an1化简即(6n1-42)d-2(n1-5)d2=
a9=a5+4da15=a5+10d(a5+4d)²=a5(a5+10d)8da5+16d²=10da516d²-2da5=02d(8d-a5)=0d=a5/8所以a9=
设公差为d(d≠0),由题意a32=a2•a6,即(a1+2d)2=(a1+d)(a1+5d),解得d=-2a1,故a1+a3+a5a2+a4+a6=3a1+6d3a1+9d=−9a1−15a1=35
因为a5=a1+4d,a9=a1+8d,a15=a1+14d且a5a9a15成等比数列所以(a1+8d)^2=(a1+4d)(a1+14d)即(a1)^2+16a1*d+64d^2=(a1)^2+18
1.若n=4时,则原数列为a1,a2,a3,a4.⑴若删去a1,则a3∧2=a2×a4,→d=0,矛盾⑵若删去a2,→a5=0矛盾⑶若删去a3→a1=d→a1/d=1⑷若删去a4→d=0矛盾综上所述,
都正确,证明过程如下(1){an}是等差数列,d为公差且不为0,a1和d均为实数,他的前n项和记作Sn,所以an=a1+(n-1)d,Sn=na1+n(n-1)d/2集合A={(an,Sn/n|n∈N
S2-S1=(an+1-a1)+(an+2-a2)+...+(a2n-an)=nd*n=d*n^2S3-S2=(a2n+1-a1)+(a2n+2-a2)+...+(a3n-a2n)=nd*n=d*n^
(1)由题意可得(a1+d)2+(a1+2d) 2=(a1+3d)2+(a1+4d)27a1+21d=7联立可得a1=-5,d=2∴an=-5+(n-1)×2=2n-7,sn=−5n+n(n
由题意,显然该等比数列的公比不会是负数,也不会是小于一的数.前者不会满足等差数列要求,后者末项趋于零,不合理.故公比大于一,故等差数列是递增的即公差大于0.又a5*a5=a3*an1即36=a3*an
首项为a1,公差为dS10=10a1+45d=110.(1)a1,a2,a4成等比数列.(a2)^2=a1*a4(a1+d)^2=a1(a1+3d).(2)通过(1)(2)得a1=d=2an=a1+(
ak=a1+(k-1)d=9d+(k-1)d=(k+8)da2k=a1+(2k-1)d=9d+(2k-1)d=(2k+8)d又a1a2k=ak^2,即9d(8+2k)d=[(8+k)d]^2k=4
设该等差数列是首项为a1,公差为dS3=3a1+3(3-1)*d/2=3a1+3dS2=2a1+2(2-1)*d/2=2a1+dS4=4a1+4(4-1)*d/2=4a1+6d又:S3²=9
在等差数列中,公差d不为0,a11+40d=a51,即a11=a51-40d因为|a11|=|a51|,即a11=-a51,或者a11=a51(不符,舍去)所以a11+a51=2*a31=0,即a31
a1a2a3成等比数列a2^2=a1a3=a3(a1+d)^2=a1+2da1^2+2a1d+d^2=a1+2d1+2d+d^2=1+2dd^2=0d=0公差不为零的等差数列错题
数列{an}是公差不为0的等差数列,设公差为d,S1,S2,S4成等比数列,则S22=S1•S4,∴( 2a1+d)2=a1•(4a1+6d),化简可得d=2a1∴a3a1=a1+2da1=
a2=a1+da3=a1+2da6=a1+5d由等比数列性质(a1+2d)^2=(a1+d)(a1+5d)a1=-1/2dq=a3/a2=3