已知是公差不为零的等差数列A1=1且A1A2A3成等比数列
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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
数列a1a2a5等比数列则有a2*a2=a1*a5a3-2d=a1a3+2d=a5a3-d=a2带入得到d=2b1+2b2+4b3+2^(n-1)bn=an(1)b1+2b2+4b3+2(n-3)bn
(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
设公差为d则a3=a1+2d=1+2da9=a1+8d=1+8d因为a1,a3,a9成等比数列所以a3²=a1*a9=a9∴(1+2d)²=1+8d∴d=0或者d=1又∵d≠0,∴
an=a1+(n-1)d=2+(n-1)da2=2+da4=2+3da8=2+7da2,a4,a8成等比数列,即a4/a2=a8/a4a4*a4=a2*a84+12d+9d^2=4+16d+7d^22
因为b1=a1²,b2=a2²;所以b1>0,q>0且q≠1({an}公差不为零)所以a1=√b1,a2=√(b1*q),a3=q*√b12a2=a1+a3->2√(b1*q)=√
令{an}公差为d,由b2^2=b1*b3得:a2^4=a1^2*a3^2两边开方得:a2^2=a1*a3或a2^2=-a1*a3当a2^2=a1*a3时,有: (a
(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
(1)由题意,设公差为d,则a1+4d=10(a1+2d)2=a1(a1+8d)∴a1+4d=104d2=4a1d∵d≠0,∴a1=2,d=2∴an=2+(n-1)×2=2n;(2)由(1)知,Sn=
1)因为an为等差数列所以a1=5-2da2=5-da5=5+2d又a1,a2,a5成等比数列所以(a2)^2=a1*a5既(5-d)^2=(5-2d)*(5+2d)又d≠0解得d=2则a1=1an=
因为a1,a2,a5成等比数列得到(a2)2=a1a5,即(a1+d)2=a1(a1+4d),化简得d(d-2a1)=0,解得d=0(舍去),d=2a1又因为a1+a2+a5>13,所以3a1+5d>
由a1+a2=a3得,a1=d,所以an=a1+(n-1)d=nd,又a1a2=a4,所以d=2,所以an=2n
(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
(1)a3=a1+2d、a6=a1+5d.(a1+2d)^2=a1(a1+5d)a1^2+4a1d+4d^2=a1^2+5a1d4a1d+4d^2=5a1d因为d0,所以4a1+4d=5a1a1=4d
(1)设等差数列{an}的公差为d(d≠0),由a1,a3,a13成等比数列,得a32=a1•a13,即(1+2d)2=1+12d得d=2或d=0(舍去).故d=2,所以an=2n-1(2)∵bn=2
解a1=1a2=1+da5=1+4da1a2a5成等比所以(1+d)^2=1*(1+4d)d^2-2d=0d=2d=0(舍)所以an=a1+(n-1)d=1+(n-1)*2=2n-1
设公差为d,则a2=1+d,a5=1+4d,则1×(1+4d)=(1+d)2,∴d=2,∴an=2n-1,故答案为:2n-1.
a1a2a3成等比数列a2^2=a1a3=a3(a1+d)^2=a1+2da1^2+2a1d+d^2=a1+2d1+2d+d^2=1+2dd^2=0d=0公差不为零的等差数列错题
设首项为a1,公差为d.由题得:a1+a5=2*4a1*a7=a₃^2则:a1+(a1+4d)=8a1(a1+6d)=(a1+2d)^2综上解得a1=2d=1所以S5=20
(I)设等差数列{an}的公差为d,由题意知d为非零常数∵a1=1,a1、a3、a9成等比数列∴a32=a1×a9,即(1+2d)2=1×(1+8d),解之得d=1(舍去0)因此,数列{an}的通项公