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在递增等差数列An中,A1=2A3是A1和A9的等比中项 ⑴...

设公差为d ∵a1=2 ∴a3=a1+2d=2+2d a9=a1+8d=2+8d 又∵a3是a1和a9的等比中项 ∴a3²=a1a9 即(2+2d)²=2(2+8d) 解得:d=0或2 ∵an是递增等差数列 ∴d=2 则an=a1+(n-1)d=2n

设数列{an}的公差为d,则∵a1+a3=8,且a4为a2和a9的等比中项,∴a1+a1+2d=8,(a1+3d)2=(a1+d)(a1+8d)解之得a1=4,d=0或a1=1,d=3当a1=4,d=0时,an=4;当a1=1,d=3时,an=3n-2.

设数列{an}的公差为d,则∵a1+a3=8,且a4为a2和a9的等比中项,∴a1+a1+2d=8,(a1+3d)2=(a1+d)(a1+8d)解之得a1=4,d=0或a1=1,d=3当a1=4,d=0时,Sn=4n;当a1=1,d=3时,Sn=n+n(n?1)2×3=3n2?n2∴{an}的前n项为Sn=4n或Sn=3n2?n2

设公差为 d,则 a2 = 2+d,a3 = 2+2d,a4 = 2+3d, 根据已知得 (2+2d)^2 = (2+d)(3+3d), 解得 d = -1(舍去) 或 2, 所以通项公式 an = 2+(n-1)d = 2n 。

(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...

∵a1=2,a1,a3,a7成等比数列∴a32=a1a7设等差数列的公差d,则(2+2d)2=2(2+6d),d>0∴d=1,an=n+1∵Sn=2n+1?2.∴b1=s1=2bn=sn-sn-1=2n+1-2-2n+2=2n(n≥2)当n=1时也适合∴bn=2n(2)∵cn=abn=2n+1∴Tn=(2+1)+(22+1)+…+(2n+1)=(2+22+23+…+2n...

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设公差为d,则a3是a1和a13等比中项,得(1+2d)2=1×(1+12d).即4d+4d2=12d,则d2=2d∵d≠0,∴解得d=2,即S10=10+10×92×2=10+90=100.故选:D.

等差数列{an}中,a1=2,a3=2+2d,a11=2+10d,因为a1、a3、a11恰好是某等比数列的前三项,所以有a32=a1a11,即(2+2d)2=2(2+10d),解得d=3,所以该等比数列的公比为82=4故选D

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