www.1862.net > (文科)已知{An}是单调递增的等差数列,首项A1=3...

(文科)已知{An}是单调递增的等差数列,首项A1=3...

(Ⅰ)设公差为d,公比为q,则a2b2=(3+d)q=12①S3+b2=3a2+b2=3(3+d)+q=20②联立①②可得,(3d+7)(d-3)=0∵{an}是单调递增的等差数列,d>0.则d=3,q=2,∴an=3+(n-1)×3=3n,bn=2n-1…(6分)(Ⅱ)bn=2n-1,cn=n?2n-1,∴Tn=c1+c2+…+cnTn=1?20+...

(Ⅰ)设{an}的公差为d,{bn}的公比为q,则a2b2=(3+d)q=12,①S3+b2=3a2+b2=3(3+d)+q=9+3d+q=20,即3d+q=11,变形可得q=11-3d,②代入①可得:(3+d)(11-d)=33+2d-3d2=12,3d2-2d-21=0,(3d+7)(d-3)=0,又由{an}是单调递增的等差数列,有...

(1)设{an}的公差为d,{bn}的公比为q,则a2b2=(3+d)q=12,①∵S3+b2=3a2+b2=3(3+d)+q=9+3d+q=20,∴3d+q=11,变形可得q=11-3d,②代入①可得:(3+d)(11-d)=33+2d-3d2=12,即3d2-2d-21=0,则(3d+7)(d-3)=0,又由{an}是单调递增的等差数列...

(I)设公差为d(d>0),则∵4S3=S6,a2+2是a1,a13的等比中项,∴4(3a1+3d)=6a1+15d(a1+d+2)2=a1(a1+12d)∴a1=1d=2或a1=?14d=?12∵d>0,∴a1=1d=2∴数列{an}的通项公式an=2n-1;(II)若存在m,k∈N*,使am+am+4=ak+2,则2m-1+2(m+4)-1=2...

∵数列{an}是单调递增的等差数列,前三项的和为12,∴3a2=12,解得a2=4,设其公差为d,则d>0.∴a1=4-d,a3=4+d,∵前三项的积为48,∴4(4-d)(4+d)=48,解得d=2或d=-2(舍去),∴a1=4-2=2,故选:B.

设等差数列的公差为d,∵a1+a2+a3=3a2=12∴a2=4∵前三项的积为48即(a2-d)a2(a2+d)=48解得d2=4∵数列{an}是单调递增的等差数列,∴d>0∴d=2故答案为2

先存着,明天一早给你解答(如果没人回答的话)

(1)因为{an}为等比数列,所以a3?a4=a2?a5=32所以a2+a5=18a2?a5=32所以a2,a5为方程 x2-18x+32=0的两根;又因为{an}为递增的等比数列,所以 a2=2,a5=16,q3=8,从而q=2,所以an=a2?qn?2=2?2n?2=2n?1;(2)由题意可知:bn=2+(n-1)d...

∵单调递增的等差数列{an}的前n项和为Sn,且S4≥10,S5≤15∴S4=4a1+4×32d≥10S5=5a1+5×42d≤15即2a1+3d≥5a1+2d≤3∴a4=a1+3d≥5?3d2+3d≥5+3d2a4=a1+3d=(a1+2d)+d≤3+d∴5+3d2≤a4≤3+d,5+3d≤6+2d,0<d≤1∴52<a4≤3+d≤3+1=4故选A.

解答 单调递增 那么d>0 a3=1 a2a4=(a3-d)(a3+d)=3/4 (1-d)(1+d)=3/4 1-d^2=3/4 d1=1/2 d2=-1/2舍弃 所以a1+2d=1 a1+1=1 a1=0

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