设sn是等比数列的前n项和 a3 2a6
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S3+S6=2S9(a1-a3*q)/(1-q)+(a1-a6*q)/(1-q)=2(a1-a*q)/(1-q)a1-a3*q+a1-a6*q=2(a1-a9*q)(a3+a6)*q=2a9*qa3+
a1+a1q>2a1q^22q^2-q-1
依据题意,有2*3a2=a1+3+a3+4=7+a1+a3=7+a1+a2+a3-a2=7+7-a2=14-a2.2*3a2=14-a26a2=14-a27a2=14.a2=2.s3=a1+a2+a3
设等差数列的等差为d,等比数列的等比是q,由a3=b3,得a4-d=b4q,又∵a4=b4,∴a4-a4q=d,∵S5-S3T4-T2=7,∴a5+a4b4+b3=a4+d+a4a4+a4q=7,即3
题目应该缺少条件,通过第二问,则应该q≠1(1)设等比数列的公比为q∵a5.a3.a4成等差数列∴2a3=a5+a4∴2a1q²=a1q^4+a1q^3即q²+q-2=0∴(q+2
设等差数列的等差为d,等比数列的等比是q则a3=b3a4-d=b4/q又∵a4=b4∴a4-d=a4/qa4-a4/q=d∵(S5-S3)/(T4-T2)=5∴(a5+a4)/(b4+b3)=(a4+
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-
不一定,当S1,S2,S3.Sn都相等时,a2,a3.an为0数列,不成等比.当S1,S2,S3.Sn公比不为1时,an=sn-s(n-1)不为0,则有a(n+1)/an=[s(n+1)-s(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
S3=a1+a2+a3=3a2=12a2=4设公差为d,则a1=a2-d=4-da3=a2+d=4+d2a1、a2、a3+1成等比数列,则a2²=(2a1)(a3+1)2(4-d)(4+d+
S3应该是等于12吧{an}是等差数列S3=a1+a2+a3=3a2=12a2=4设公差为da1=4-da3=4+d2a1,a2,a3+1成等比数列(a2)^2=2a1·(a3+1)4^2=2(4-d
设公比为q,则q>0a3=a2+4a1q^2=a1q+4a1=2代入,整理,得q^2-q-2=0(q+1)(q-2)=0q=-1(舍去)或q=2Sn=a1(q^n-1)/(q-1)=2×(2^n-1)
a(1)=s(1)=2a(1)-2,a(1)=2,s(n)=2a(n)-2^n,s(n+1)=2a(n+1)-2^(n+1),a(n+1)=s(n+1)-s(n)=2a(n+1)-2^(n+1)-2a
(Ⅰ)当q=1时,S3=3a1,S9=9a1,S6=6a1,∵2S9≠S3+S6,∴S3,S9,S6不成等差数列,与已知矛盾,∴q≠1.(2分)由2S9=S3+S6得:2•a1(1−q9)1−q=a1
a3=a1*q^2;a9=a1*q^8;a6=a1*q^5;因为a3,a9,a6是等差数列,所以,2a9=a3+a6.化简,2q^9=q^3+q^6.s3+s6=a1*(1-q^3)/(1-q)+a1
(1)令n=1,得a1=-1.Sn=2an+n,S(n+1)=2a(n+1)+n+1.两式相减,得a(n+1)=2a(n+1)-2an+1.整理得a(n+1)-1=2(an-1),a1-1=-2.综上
令an公比为q,则由2a3-a4=0得,2q²-q³=0,由于q≠0,所以q=2S5/a1=(a5+a4+a3+a2+a1)/a1=q^4+q³+q²+q+1=