公差不为零的等差数列an的前23项和等于其前10项和,a8 ak=0则正整数
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1.因为等差数列AN的公差d不等于0,a1=2,s9=36,所以36=9*2+1/2*9*8d所以d=1/2所以a3=3,a9=6,由a3,a9,am成等比数列则a9的平方=a3*am,的am=12又
(Ⅰ)设公差为d,由条件得5a1+5×42d=30(a1+2d)2=a1(a1+8d),得a1=d=2.∴an=2n,Sn=2n+n(n-1)×22=n2+n;(Ⅱ)∵1Sn+an+2=1n2+n+2
(1)a3=a1+2d,a7=a1+6d,所以a1*a7=a3*a3,即a1*(a1+6d)=(a1+2d)*(a1+2d)解得d=1(2)Sn=(1/2)n^2+(3/2)n,又a3=a1+2d=4
(本题满分14分)(1)∵Sn=2an-2,∴当=1时,a1=2a1-2,解得a1=2;当n=2时,S2=2+a2=2a2-2,解得a2=4;当n=3时,s3=a1+a2+a3=2a3-2,解得a3=
(1)根据题意,设公差为d则a3=a1+2d=2d+1a9=a1+8d=8d+1有(2d+1)^2=8d+1d=1故通项:an=n(2)根据题意,设公比为q则b2=qb3=q^2有q-0.5q^2=0
a1*a5=a3*a4a1*(a1+4d)=(a1+2d)(a1+3d)d=2Sn最小即an
2^n等比数列啊Sn=2+2^2+.+2^n2Sn=2^2+.+2^(n+1)2Sn-Sn=2^(n+1)-2Sn=2^(n+1)-2
a9=a5+4da15=a5+10d(a5+4d)²=a5(a5+10d)8da5+16d²=10da516d²-2da5=02d(8d-a5)=0d=a5/8所以a9=
a2^2+a3^2=a4^2+a5^2a2^2+(a2+d)^2=(a2+2d)^2+(a2+3d)^2解得d=2a2/3Sn7=7a1+3d=1解得d=2/7a1=1/7an=1/7+(n-1)2/
(1)∵数列{an}是公差不为零的等差数列,a1=2,且a2,a4,a8成等比数列,∴(2+3d)2=(2+d)(2+7d),解得d=2,∴an=2n.(2)∵an=2n,∴3an=32n=9n,此数
(I)由题意可得,4a1+6d=10(a1+2d)2=(a1+d)(a1+6d)∵d≠0∴a1=−2d=3∴an=3n-5(II)∵bn=2an=23n-5=14•8n−1∴数列{an}是以14为首项
用求和公式,求解二元二次方程组.
(1)由题意可得(a1+d)2+(a1+2d) 2=(a1+3d)2+(a1+4d)27a1+21d=7联立可得a1=-5,d=2∴an=-5+(n-1)×2=2n-7,sn=−5n+n(n
由a1+a3=4知a1+(a1+2d)=4即a1+d=2,又a2,a3,a5成等比数列得到a32=a2a5即(a1+2d)2=(a1+d)(a1+4d),a12+4da1+4d2=a12+5da1+4
首项为a1,公差为dS10=10a1+45d=110.(1)a1,a2,a4成等比数列.(a2)^2=a1*a4(a1+d)^2=a1(a1+3d).(2)通过(1)(2)得a1=d=2an=a1+(
设该等差数列是首项为a1,公差为dS3=3a1+3(3-1)*d/2=3a1+3dS2=2a1+2(2-1)*d/2=2a1+dS4=4a1+4(4-1)*d/2=4a1+6d又:S3²=9
s1=a1s2=2a1+ds4=4a1+6d因为s1,s2,s4成等比数列所以(s2)²=s1×s4(2a1+d)²=a1(4a1+6d)4a1²+4a1d+d²
数列{an}是公差不为0的等差数列,设公差为d,S1,S2,S4成等比数列,则S22=S1•S4,∴( 2a1+d)2=a1•(4a1+6d),化简可得d=2a1∴a3a1=a1+2da1=
a2=a1+da3=a1+2da6=a1+5d由等比数列性质(a1+2d)^2=(a1+d)(a1+5d)a1=-1/2dq=a3/a2=3