设等差数列an不为0a1=qd
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摘下来的:邦你分析下,先证必要性(如果是等差,则...);(所以可以)设数列an的公差为d,若d=0,则所述等式显然成立.若d≠0,则==.再用数学归纳法证明充分性:所述的等式对一切n∈N都成立,首先
(1)令通项公式:an=a1+(n-1)da2=a1+da4=a1+3dS10=5(2a1+9d)=110由题意:a2^2=a1*a4即(a1+d)^2=a1*(a1+3d)由题意:a1=d=2所以通
由AK是A1与A2k的等比中项,得得(AK)^2=A1*A2K因为A1=9d所以AK=8+KdA2K=8+2Kd所以(8+Kd)^2=9d*(8+2Kd)(K-4)*(k+2)=0因为K>0所以K=4
a3²=a1a13(a1+2d)²+a1(a1+12d)a1=1所以1+4d+4d²=1+12d4d²-8d=0所以d=2所以an=2n-1bn=2^)2n-1
设公差da2=a1+da4=a1+3da10=a1+9dS10=(a1+a10)*10/2=5(2a1+9d)=1102a1+9d=22a1.a2.a4为等比数列a2*a2=a1*a4(a1+d)^2
(1)由题设可知公差d≠0,由a1=1且a1,a3,a9成等比数列,得:(1+2d)2=1+8d,解得d=1或d=0(舍去),故{an}的通项an=n.(2)∵bn=2 an=2n,∴数列{
a1=9dak=a1+(k-1)*d=9d+(k-1)*da2k=a1+(2k-1)*d=9d+(2k-1)*dak^2=a1*a2k化简后可求出k=4
由题意可得:a3=2+2d,a6=2+5d由a1,a3,a6成等比数列所以(2+2d)^2=2(2+5d)又d不为0解得d=1/2由等差数列Sn=a1*n+n(n-1)d/2可得:Sn=2n+n(n-
a3=a1+d=2+2da6=a1+5d=2+5d等比数列,所以(2+2d)²=2*(2+5d)4+8d+4d²=4+10d4d²=2dd不等于0d=1/2an=2+1/
a1,a3,a6成等比数列a3²=a1a6(a1+2d)²=a1(a1+5d)a1²+4a1d+4d²=a1²+5a1da1d=4d²d≠0
设a3=a,公差为d则a2=a-d,a6=a+3d成等比数列,即(a2)*(a6)=(a3)*(a3)代入得出3d=2a.即d=2/3a所以公比为a3/a2=a/(a-d)=a/(1/3a)=3即公比
公差为da3=2+2da6=2+5d成等比数列,则a3^2=a1*a6(2+2d)^2=2(2+5d)4d^2+8d+4=4+10d4d^2-2d=02d(2d-1)=0d=1/2(因为d不为0)an
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
a1=9d则ak=9d+(k-1)d,a2k=9d+(2k-1)d因为ak为a1和ak的等比中项则有ak的平方等于a1乘以a2k即{9d+(k-1)d}^2=9d{9d+(2k-1)d}化简消去d得:
公差为da3=2+2da6=2+5d成等比数列,则a3^2=a1*a6(2+2d)^2=2(2+5d)4d^2+8d+4=4+10d4d^2-2d=02d(2d-1)=0d=1/2(因为d不为0)an
S1/a1=1S2/a2-S1/a1=(2+d)/(1+d)-1=d/(1+d)S3/a3-S1/a1==(3+3d)/(1+2d)-1=(2+d)/(1+2d)2*d/(1+d)=(2+d)/(1+
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
(1)设等差数列{an}的公差为d,由a22=a1a4,…(1分)得(a1+d)2=a1(a1+3d)…(2分)∵d≠0,∴d=a,∴an=na1,Sn=an(n+1)2.(2)∵1Sn=2a(1n−
因为ak是a1与a2k的等比中项,则ak2=a1a2k,[9d+(k-1)d]2=9d•[9d+(2k-1)d],又d≠0,则k2-2k-8=0,k=4或k=-2(舍去).故选B.
A1=1,A2=1+d,A8=1+7d;B1=1,B2=1*q,B3=1*q^2=>1+d=q;1+7d=q^2=>d=5,q=6,A2=B2=6,A8=B3=36S(Bn)=A1(1-q^n)/(1