在公差不为0的等差数列an中,a1a2为方程x²-a3x a4=0
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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……
设An=A1+(n-1)d则A2=A1+dA4=A1+3d因为A2是A1与A4等比中项故(A2)²=A1A4即(A1+d)²=A1(A1+3d)d²=A1d因为d不为0,
(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
2a3-a7²+2a11=0a3+a11=2a74a7-a7²=0a7=4b7=4b6b8=b7²=49
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)a5=6a3=3am=1.5m-1.5=12m=9(2)a3=2an=2n-4q=6/2=3a(nt)/a(n(t-1))=3*a(n(t-1))a(nt)=2*3^(n+1)a(nt)=2nt
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+
a2=a1+da4=a1+3da6=a1+5da2,a4-2,a6成等【比】数列(a1+3d-2)^2=(a1+d)(a1+5d)(3d-1)^2=(1+d)(1+5d)9d^2-6d+1=5d^2+
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
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
公差是:m^2d证明:S2m-Sm=(a1+a2+……+a2m)-(a1+a2+……+am)=a(m+1)+a(m+2)+……+a2m同理S3m-S2m=a(2m+1)+a(2m+2)+……+a3m所
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
a2,a3,a6组成等比数列的连续三项∴a3的平方=a2a6(a1+2d)²=(a1+d)(a1+5d)化简得d=-2a1q=a3/a2=(a1+2d)/(a1+d)=(-3a1)/(-a1
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