已知数列an为公差不为0的等差数列Sn为前n项和a5和a7的等差中项
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数列an满足条件:A1=1,A2=r(r>0)数列{an+an+1}是公差为d的等差数,令bn=an+an+1即首项b1=a1+a2=1+rb3=a3+a4=b1+2d=1+r+2db5=a5+a6=
即对任意n∈N,(a+n)/(a+n-1)≥(a+8)/(a+7)两边同减1:1/(a+n-1)≥1/(a+7)此不等式可分三种情况:(1)a+7≥a+n-1〉0显然n≥8时不成立(2)0〉a+n-1
a1=a2-d,a5=a2+3d所以a2a2=(a2-d)(a2+3d)得2da2=3dd即a2=3d/2所以a1=a2-d=1d/2=1得出d=2公差=2,首项=1,后面你会的即a10=19故S10
a2=a1+da4=a1+3da6=a1+5da2,a4-2,a6成等【比】数列(a1+3d-2)^2=(a1+d)(a1+5d)(3d-1)^2=(1+d)(1+5d)9d^2-6d+1=5d^2+
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
设公差为d,则d≠0a1,a3,a9成等比数列,则a3²=a1·a9(a1+2d)²=a1(a1+8d)a1=1代入,整理,得d²-a1d=0d(d-a1)=0d≠0,因
(1)当n=4时有a1,a2,a3,a4.将此数列删去某一项得到的数列(按照原来的顺序)是等比数列.如果删去a1,或a4,则等于有3个项既是等差又是等比.可以证明在公差不等于零的情况下不成立(a-d)
(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.设数列{an}的公差是d,则a(n+1)cosA+an*sinA=(an+d)*cosA+an*sinA=1即(cosA+sinA)*an=1-dcosA若cosA+sinA不等于0,则an=(1
因为a(k1),a(k2),…,a(kn)恰为等比数列,又k1=1,k2=5,k3=17所以a5的平方=a1乘以a17又因为数列{an}为等差数列且公差d≠0所以a5=a1+4da17=a1+16d所
设an=a+d*(n-1)1.a3+a10=a+2d+a+9d=2a+11d=152.a3*a7=a4*a4(a+2d)(a+6d)=(a+3d)^2a=-1.5d联立1与2,求得d=15/8a=-4
S1/a1=1S2/a2-S1/a1=(2+d)/(1+d)-1=d/(1+d)S3/a3-S1/a1==(3+3d)/(1+2d)-1=(2+d)/(1+2d)2*d/(1+d)=(2+d)/(1+
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
(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
a5=1+4da2=1+d1+4d=(1+d)^2d^2-2d=0d≠0d=2an=1+2(n-1)=2n-1
解∵a1,a2,a5是等比数列∴a2²=a1a5由a1=1∴a2²=a5∴(1+d)²=1+4d∴1+2d+d²=1+4d即d²-2d=0∴d=0或d
设数列{an}是公差为d,且d≠0,因为a5,a10,a20三项成等比数列,所以(a1+9d)2=(a1+4d)(a1+19d),整理得5a1d=5d2,解得d=a1,则公比q=a10a5=a1+9d
a1+a2+...+an=(1/2)(an²+an)a1+a2+...+a(n-1)=(1/2)(a(n-1)²+a(n-1))两式相减得an=(1/2)(an²+an)