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已知{An}是等比数列,A1=2,A5=1/5,则A1A2+A2A...

1、设差为n, 比为q 则a2=2+n=2q a5=2+4n=2q^2 根据题意可得:n=2q-2 得 2+8q-8=2q^2 q=1或3 则n=0或4 根据题意n=4 An=2+4(n-1) 等差数列通项公式:an=a1+(n-1)*d,n为正整数 等比数列通项公式:an=a1q^n-1,公比q≠0,等比数列a1≠ 0。其中an中...

数学知识都还给老师了哈哈,回答你第一问。 a1+a2=2a1+d=1 a2,a3,a5成等比数劫,a5/a3=a3/a2,即(a1+4d)/(a1+2d)=(a1+2d)/(a1+d),化简这个等式,得到a1d=0,因为d不等于0,所以a1=0。 2a1+d=1,得知d=1 所以数列{an}的通项公式为:an=a1+(n-1...

由题意等比数列{an}a>0,n=1,2,…,当n>1时,log2a1+log2a3+log2a5+…+log2a2n+1=log2a1a3a5…a2n+1.又a5?a2n-5=22n(n≥3)∴a1a3a5…a2n+1=2(n+1)2∴log2a1+log2a3+log2a5+…+log2a2n+1=log22(n+1)2=(n+1)2故选:C.

(1)由等差数列{an}是递增数列,可设{an}的公差为d(d>0),∵a1,a2,a5成等比数列,S5=a32,∴a22=a1a55a3=a32,解得a1=1d=2,∴an=2n-1.(2)假设存在正整数m,l,使数列am,am+l,am+2l为等比数列,则am+l2=amam+2l,而an=2n-1,∴[2(m+...

∵等差数列{an}的公差和首项都不等于0,且a2,a4,a8成等比数列,∴a42=a2a8,∴(a1+3d)2=(a1+d)(a1+7d),∴d2=a1d,∵d≠0,∴d=a1,∴a1+a5+a9a2+a3=15a15a1=3.故选:B.

由a5?a2n-5=an2=22n,且an>0,解得an=2n,则log2a1+log2a2+log2a3+…+log2a2n-1=log(a1a2n?1)?(a2a2n?2) …an 2=log2n22=n2.故答案为:n2

a1*(1+q+q^2)=1 a4*(1+q+q^2)=-2 所以a4/a1=-2 q=-2^(1/3)

a1*a2*a3=a2^3=1,a2=1 a4*a5*a6=a5^3=8,a5=2 a5/a2=q^3=2 a8=a5*q^3=2*2=4

设公比为q,则a(n+10)=an×q^10 ∴a11+a12+a13+......a20=﹙a1+a2+a3+......+a10﹚×q^10 a21+a22+a23+......a30=﹙a11+a12+a13+......a20﹚×q^10=﹙a1+a2+a3+......+a10﹚×q^20 ∴a11+a12+a13+......a30=﹙a1+a2+a3+......+a10﹚×﹙q^10+q^20) ∴...

a2=b2 a8=b4 a32=b6 1+d=b2,......................(1) 1+7d=b1q^3=b2*q^2..............(2) 1+31d=b1q^5=b2*q^4............(3) (2)-(1)得 6d=b2(q^2-1)................................(4) (3)-(1)得 30d=b2(q^4-1)............................

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