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已知公差不为0的等差数列{An}满足A1,A3,A4成等比...

我晕ヽ(・_・;)ノ写错一步

(Ⅰ)设公差为d,∵公差不为零的等差数列{an},满足a3=5且a1,a2,a4成等比数列.∴a1+2d=5(a1+d)2=a1(a1+3d)a1≠0,解得:a1=53,d=53,得an=53n(n∈N*)(Ⅱ)由题意an=53n,∴bn=153n?53(n+1)=925(1n?1n+1),∴Tn=925[(1?12)+(12?13)+…+(1n?1...

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...

解答:(1)解:设公差为d≠0,∵a3=6,且a1,a2,a4成等比数列,∴a1+2d=6,且(a1+d)2=a1?(a1+3d),解得a1=2,d=2.∴数列{an}的通项公式为an=2+(n-1)×2=2n;∵bn+1=2bn+1,∴bn+1+1=2(bn+1),∵b1=3,∴数列{bn+1}是以4为首项,2为公比的等比...

解: (1) 设公差为d,d≠0 a2,a3,a5成等比数列,则 a3²=a2·a5 (a1+2d)²=(a1+d)(a1+4d) 整理,得 a1d=0 d≠0,因此只有a1=0 a1+a2=1 a2=1-a1=1-0=1 d=a2-a1=1-0=1 数列{an}是以0为首项,1为公差的等差数列 an=0+1×(n-1)=n-1 数列{an}的...

a3=a1+2d,a6=a1+5d,a7=a1+6d 由题,(a1+5d)^2=(a1+2d)(a1+6d) 化简得a1=-13d/2 则a4=a1+3d=-7d/2,a6=a1+5d=-3d/2 a4/a6=7/3

解: 设公差为d,则d≠0。 a4是a3、a7的等比中项,则 a4²=a3·a7 (a3+d)²=a3·(a3+4d) 整理,得2a3d-d²=0 d(2a3-d)=0 d=0(舍去)或2a3-d=0 a3=d/2 a1=a3-2d=d/2 -2d=(-3/2)d S8=8a1+8×7d/2=8·(-3/2)d+28d=16d=32 d=2 a1=(-3/2)d=(-3/...

∵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.

(1)在等差数列{an}中,设其公差为d,(d≠0),∵a1a5=a22,a3=5,∴(a3-2d)(a3+2d)=a22,即(5-2d)(5+2d)=(5-d)2,…2分化简得5d2-10d=0,∴d=2…4分∴an=a3+(n-3)d=5+2(n-3)=2n-1…7分(2)∵b1+2b2+22b3+…+2n-1bn=an,①∴b1+2b2+22b3+…+...

a1/a3=a3/a4 a1/(a1+4)=(a1+4)/(a1+6) a1²+6a1=a1²+8a1+16 2a1=-16 a1=-8 a9=-8+2X(9-1)=8 S9=(-8+8)x9÷2=0

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