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已知数列{An}是等差数列,其前n项和为Sn,若A1A2A3...

∵数列{an}是等差数列,∴S1=a1,S3=3a2,S5=5a3,又∵3S1S3+15S3S5+5S5S1=35∴1a1a2+1a2a3+1a1a3=35又∵a1a2a3=15∴a315+a115+a215=a25=35,∴a2=3故选C

设公差为x,则有 a1=1/2 s2=a3 2a1+x=a3 a3-a1-a1=2x-a1 a3-a1-a1=x 代入得x-a1=0 x=1/2 则公差为1/2 所以a2=a1+x=1 所以Sn=na1+n(n-1)x/2=n1/2+n(n-1)/4=n(n-1+2)/4=n(n+1)/4=(n^2+n)/4 那两楼对Sn的计算都是错误的,我验算了,我这答案应该是...

(Ⅰ)设等差数列{an}的公差等于d,则由题意可得2a1+2d=82a1+4d=12,解得 a1=2,d=2.∴{an}的通项公式 an =2+(n-1)2=2n.(Ⅱ) 由(Ⅰ)可得 {an}的前n项和为Sn =n(a1+an)2=n(n+1).∵若a1,ak,Sk+2成等比数列,∴ak2=a1 Sk+2 ,∴4k2 =2(k+2...

(1){an}为等差数列,所以a1+a4=a2+a3=14,又a2a3=45,所以a2,a3是方程x2-14x+45=0的两实根,公差d>0,∴a2<a3∴a2=5,a3=9∴a1+d=5a1+2d=9?a1=1d=4所以an=4n-3(2)由(1)知sn=2n2-n,所以bn=snn+c=2n2?nn+c∴b1=11+c,b2=62+c,b3=1...

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(本小题满分12分)解:(Ⅰ)设数列{an}的公差为d(d≠0),∵S3=a4+2,∴3a1+3×2×d2=a1+3d+2.①…(3分)又∵a1,a2-1,a3-1成等比数列,∴a1(a1+2d?1)=(a1+d?1)2.②…(5分)由①②解得a1=1,d=2.…(6分)∴an=a1+(n-1)d=2n-1.…(7分)(Ⅱ)∵1anan...

设等比数列{an}的公比为q,则可得a1q?a1q2=2a1,即a4=a1q3=2又a4与2a7的等差中项为54,所以a4+2a7=52,即2+2×2q3=52,解之可得q=12,故a1=16故S6=16(1?126)1?12=632故答案为:632

1)因为an为等差,故a1+a3=2*a2;所以S3=a1+a2+a3=a2+(a1+a3)=3*a2=12; 故:a2=4;设等差数列{an}的等差为d(d>0);则a1=a2-d=4-d;a3=a2+d=4+d; 所以数组2*a1,a2,a3+1为等比(等价于)数组2*(4-d),4,4+d+1为等比;化简为:2*(4-d),4,5+d 又等比性质中...

(Ⅰ)∵a1,23a2,13a3依次成等差数列,∴43a2=a1+13a3,即:4a2=3a1+a3.设等比数列{an}公比为q,则4a1q=3a1+a1q2,∴q2-4q+3=0.∴q=1(舍去),或q=3.又S3=a1+a1q+a1q2=13a1=39,故a1=3,∴an=3n. (Ⅱ) 当n≥2时,bn=3n?(13+132+…+13n?1)...

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