若在公差不为0的等差数列中a1 a3=8
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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
(1)a3=a1+2d,a7=a1+6d,所以a1*a7=a3*a3,即a1*(a1+6d)=(a1+2d)*(a1+2d)解得d=1(2)Sn=(1/2)n^2+(3/2)n,又a3=a1+2d=4
a1+q^2*a1=2*q*a1解得q=1不存在满足条件的答案……你检查题目是不是有问题……
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
(1)由a2=b2a8=b3a1=b1=1得1+d=q1+7d=q2(3分)∴(1+d)2=1+7d,即,d2=5d,又∵d≠0,∴d=5,从而q=6(6分)(2)∵an=a1+(n-1)d=5n-4
易知(a2+d)^2=a2*(a2+4d)得:d=2a2所以(a1+a3+a5)/(a2+a4+a6)=(a2-d+a2+d+a2+3d)/(a2+a2+2d+a2+4d)=(a2+a2+a2+6a2
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:
你的答案似乎不对,因为我做过这道题三遍.1.a1,a3,a5成等比则:a3^2=a1*a5又a1,a3,a5是等差数列{an}中的项则:a3=a1+2da5=a1+4d则有:(a1+2d)^2=a1(
(1)设等差数列{an}的公差为d(d≠0),由a1,a3,a13成等比数列,得a32=a1•a13,即(1+2d)2=1+12d得d=2或d=0(舍去).故d=2,所以an=2n-1(2)∵bn=2
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
(Ⅰ)设等差数列{an}的公差为d(d≠0),由题意得a22=a1a4,即(a1+d)2=a1(a1+3d),∴(2+d)2=2(2+3d),解得 d=2,或d=0(舍),∴an=a1+(n
设a2=b2=x则a5=4x-3b3=x^2所以4x-3=x^2解得x=1(舍去,因为公差不为0)或者3所以(1)an=2n-1bn=3^(n-1)(2)S(bn)=(3^n-1)/2(3)若成立则2
假设等差数列公差为d,等比数列公比为q,则由题意可得:a2=a1+d=1+db2=b1*q=qa6=a1+5d=1+5db3=b1*q^2=q^2(注:代表q平方)由a2=b2和a6=b3,得1+d=
(1)由已知a1=b1=1,a2=b2,a8=b3得1+d=q,1+7d=q^2解方程组可得出d=5,q=6或者d=0,q=1(不符舍去)∴d=5,q=6则通项公式为an=1+(n-1)*5=5n-4
a1=b1a3=b3a2=(a1+a3)/2b2=根号b1b3=根号a1a3因为a1>0,a3>0,a1≠a3所以a1+a3>2根号a1a3(a1+a3)/2>根号a1a3所以a2>b2a5-a3=a
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