公差不为0的等差数列an的三项a1,a4,a16成等比数列,则

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

设公差da2=a1+da4=a1+3da10=a1+9dS10=(a1+a10)*10/2=5(2a1+9d)=1102a1+9d=22a1.a2.a4为等比数列a2*a2=a1*a4(a1+d)^2

（Ⅰ）由题知a\x0523=a1a7,设等差数列{an}的公差为d,则(a1+2d)2=a1（a1+6d）,a1d=2d2,∵d≠0∴a1=2d．…（1分）又∵a2=3,∴a1+d=3a1=2,d=1

由题意可得：a3=2+2d,a6=2+5d由a1,a3,a6成等比数列所以(2+2d)^2=2(2+5d)又d不为0解得d=1/2由等差数列Sn=a1*n+n(n-1)d/2可得：Sn=2n+n(n-

1)设公差为d已知(a4)^2=a2*a5则(a1+3d)^2=(a1+d)(a1+4d)a1*d+5d^2=05d=-a1=10d=2故通项公式an=-10+2(n-1)=2n-122)bn=a^[

a3=a1+d=2+2da6=a1+5d=2+5d等比数列,所以(2+2d)²=2*(2+5d)4+8d+4d²=4+10d4d²=2dd不等于0d=1/2an=2+1/

a1,a3,a6成等比数列a3²=a1a6(a1+2d)²=a1(a1+5d)a1²+4a1d+4d²=a1²+5a1da1d=4d²d≠0

设a3=a,公差为d则a2=a-d,a6=a+3d成等比数列,即(a2)*(a6)=(a3)*(a3)代入得出3d=2a.即d=2/3a所以公比为a3/a2=a/(a-d)=a/(1/3a)=3即公比

公差为da3=2+2da6=2+5d成等比数列,则a3^2=a1*a6(2+2d)^2=2(2+5d)4d^2+8d+4=4+10d4d^2-2d=02d(2d-1)=0d=1/2(因为d不为0)an

设公差为d（d≠0），由题意a32=a2•a6，即（a1+2d）2=（a1+d）（a1+5d），解得d=-2a1，故a1+a3+a5a2+a4+a6=3a1+6d3a1+9d=−9a1−15a1=35

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

设a1=a,则a7=a+6da10=a+9da15=a+14d所以(a+9d)^2=(a+6d)(a+14d)a^2+18ad+81d^2=a^2+20ad+84d^22ad+3d^2=0d≠02a=

因为{an}为等差数列，由a1，a3，a4成等比关系，得到a32=a1a4即（a1+2d）2=a1（a1+3d），化简得d（a1+4d）=0由d≠0得到a1+4d=0，所以a1=-4d即a5=0，则S

因为数列{an}是公差不为零的等差数列,所以a7=a1+6d,a10=a1+9d,a15=a1+14d,又因为a7,a10,a15是等比数列{bn}的连续三项,所以(a1+6d)(a1+14d)=(a

先做个mark,回头再做给你看.----------------------------------------将{an}分拆成{bt}、{ct}数列排列如下：{bt}:a1,a3,a5,a7,a9,

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+

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+

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

设an=a1+(n-1)d则a2=a1+da3=a1+2da4=a1+3da7=a1+6d因为等差数列{an}的前四项和为10所以,a1+a2+a3+a4=10即4a1+6d=10.①又因a2,a3,

a2,a3,a6组成等比数列的连续三项∴a3的平方=a2a6(a1+2d)²=（a1+d)(a1+5d)化简得d=-2a1q=a3/a2=(a1+2d)/(a1+d)=(-3a1)/(-a1