已知公差不为零s5=25
来源:学生作业帮助网 编辑:作业帮 时间:2024/10/04 05:20:24
a1=25、a11=25+10d、a13=25+12d则:a11²=(a1)×(a13)(25+10d)²=25×(25+12d)得:d=-2则:a(n)=-2n+27数列a1、a
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
(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=
a9=a5+4da15=a5+10d(a5+4d)²=a5(a5+10d)8da5+16d²=10da516d²-2da5=02d(8d-a5)=0d=a5/8所以a9=
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=
设等差数列公差为d.S5=5a1+10d=5(a1+2d)=5a3=70a3=14a2、a7、a22成等比数列,则a7²=a2×a22(a3+4d)²=(a3-d)(a3+19d)
(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
S8=S5+a6+a7+a8因为a6+a8=2a7所以,S8=S5+3a7已知S8=S5+45所以,a7=15又已知a4,a7,a12成等比数列,则:a4*a12=a7^2=225===>(a7-3d
(1)因为a4,a5,a8成等比数列,所以a52=a4a8.设数列{an}的公差为d,则(3+3d)2=(3+2d)(3+6d)化简整理得d2+2d=0.∵d≠0,∴d=-2.于是an=a2+(n-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
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