公差不为零的等差数列其前23项和等于

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

(1)a4=a1+3d,a5=a1+4d因为a1、a4、a5成等比数列那么a4^2=a1*a5即(a1+3d)^2=a1*(a1+4d)所以2a1*d+9d^2=0因为d≠0所以2a1+9d=0①又S

（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

设该等差数列首项a1,公差d则S1=a1S2=2a1+dS4=4a1+6d要成等比(2a1+d)^2=a1(4a1+6d)即4a1^2+4a1d+d^2=4a1^2+6a1d即d=2a1所以S1=a1

a1*a5=a3*a4a1*(a1+4d)=(a1+2d)(a1+3d)d=2Sn最小即an

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-

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)求数列an的前n项和Sn(2)试求所有的正整数m,使得am*a(m+1)/a(m+2)为数列an中的项设an=a+(n-1)dd不等于0（a2）^2+(a3)^2=（a4）^2+(a5)^2即（

（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

（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

1)设首项为a1,公差为d,a2^2+a3^2=a4^2+a5^2(a5+a3)*(a5-a3)+(a4+a2)(a4-a2)=0由于a5-a3=a4-a2=2d≠0,所以a2+a3+a4+a5=0则

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

公差为dd不为零S7=(a1+a7)*7/2=a4*7=7所以a4=1;a2^2+a3^2=a4^2+a5^2a5^2-a3^2=a2^2-a4^22d(a5+a3)=-2d(a2+a4)（a5+a3

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+(

(1)设an=a1+(n-1)dd不等于0（a2）^2+(a3)^2=（a4）^2+(a5)^2即（a1+d）^2+(a1+2d)^2=（a1+3d）^2+(a+41d)^2解得a1=-5d/2再由S

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=

(1)a2²+a3²=a4²+a5²a2²-a5²=a4²-a3²(a2+a5)*(a2-a5)=(a4+a3)*(a4