已知公差不为0的等差数列 的前n项和为Sn,若S4=20,且a1 a3 a4

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

Sn=d/2n^2+(a1-d/2)n,(1).Sn+Sm=d/2(n^2+m^2)+(a1-d/2)(n+m)>=d/2(n^2+m^2+2nm)/2+(a1-d/2)(n+m)=d/22p^2+(

（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

已知﹛an﹜是首项为-16,公差不为0的等差数列,其前n项和为Sn,且a1,a5,a4成等比数列,求﹛an﹜的公差d.设a1=-16a5=-16+4da4=-16+3d所以,a5²=a1*a

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-

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

已知{an}是首项为19,公差为-2的等差数列,Sn为｛an｝的前n项和,An=19+(n-1)×(-2)=-2n+21和=(首项+末项)×项数÷2即：Sn=(19+an)×n÷2=(-2n+40)n

都正确,证明过程如下(1){an}是等差数列,d为公差且不为0,a1和d均为实数,他的前n项和记作Sn,所以an=a1+(n-1)d,Sn=na1+n(n-1)d/2集合A={（an,Sn/n|n∈N

（Ⅰ）设公差为d，且d≠0，∵S3=a4+6，且a1，a4，a13成等比数列∴3a1+3d=a1+3d+6，（a1+3d）2=a1（a1+12d）∴a1=3，d=2∴an=3+2（n-1）=2n+1；

等差数列{an}公差为d(d≠0),前n项和Sn=na1+n(n-1)d/2∴xn=Sn/n=a1+(n-1)d/2∴{xn}为等差数列,首项为a1公差为d/2∴｛Xn｝的前n项和Tn=n[2a1+(

1/2(3/2-1/(1+n)-1/(2+n))再问：过程谢谢再答：An我就直接当成=2n+1了。1/Sn=1/n(2+n)=1/2(1/n-1/(2+n)),令1/Sn前n项和为Dn,则Dn=1/2

（1）由题意可得an=19+（n-1）×（-2）=-2n+21，a2=17∴Sn＝−n2+20n．…（6分）（2）由题意可得，bn＝3n−1由等比数列的求和公式可得，Tn＝1−3n1−3=3n−12…

由题意得a1>=0a2>0...an>0d>0sn=na1+(n-1)dsm=ma1+(m-1)dsp=pa1+(p-1)d由1/sn+1/sm>=2/sp得sp(sn+sm)>=2sn*smspsn

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+

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

（1）设等差数列{an}的公差为d，由a22＝a1a4，…（1分）得(a1+d)2＝a1(a1+3d)…（2分）∵d≠0，∴d=a，∴an=na1，Sn＝an(n+1)2．（2）∵1Sn＝2a(1n−

1/a1/(a+2d)=1/(a+d)^2a(a+2d)=(a+d)^2a^2+2ad=a^2+2ad+d^2d^2=0d=0哪儿写错了吧?再问：是1/a1,1/a2,1/a4成等比数列再答：a2^2

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,