等差数列an中,设sn为其前n项和,且a1大于0,s3等于s11
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(1)由已知,n,an,Sn成等差数列,所以Sn=2an-n,Sn-1=2an-1-(n-1),(n≥2)两式相减得an=Sn-Sn-1=2an-2an-1-1,即an=2an-1+1,两边加上1,得
你数列当中的第五个元素
设第一项为:a1,公差为:d1、S15>0可得到a1>-7d2、a8+a9
数列{Sn/n}构成一个公差为2的等差数列,∴Sn/n=2n,∴Sn=2n^2,∴a3=S3-S2=18-8=10.
2Sn=an+an^22Sn-1=an-1+an-1^2两式相减:2an=an^2+an-(an-1+an-1^2)an^2-an-(an-1+an-1^2)=0(an-(an-1+1))(an+an
Sm=Snma1+m(m-1)d/2=na1+n(n-1)d/2(m-n)a1+(m²-m-n²+n)d/2=0(m-n)a1+[(m+n)(m-n)-(m-n)]d/2=0a1(
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-
a(n)=a+(n-1)d,a>0,d>0.s(n)=na+n(n-1)d/2.2s(1)=2a(1)=2a=a(1)a(n+1)=a(a+d),0=a(a+d)-2a=a(a+d-2).a+d=2.
等差数列a2+a3-a1+a4a2*a3=45a2+a3=14d>0所以a2=5a3=9d=a3-a2=4a2=a1+da1=1an=a1+(n-1)d=4n-3Sn=na1+n(n-1)d/2=n+
2sn=(an)^2+an,2(sn+1)=(an+1)^2+(an+1)作差((sn+1)-(sn)=an+1)则((an+1)-an-1)((an+1)+an)=0因为数列{an}的各项都是正数所
Sn=a1(1-q^n)/(1-q)Sn+1=a1[1-q^(n+1)]/(1-q)Sn+2=a1[1-q^(n+2)]/(1-q)2Sn+2=Sn+Sn+1a1[1-q^(n+1)]/(1-q)+a
设公差为d,则:a1=2-3da2=2-2da3=2-da4=2a5=2+d∵S5=20∴10-5d=20d=-2∴an=10-2n以上希望对你有所帮助
Sn=a1n+n(n-1)d/2S4=4a1+6d=-62S6=6a1+15d=-75a1=-20,d=3an=a1+(n-1)d=3n-23当n<8时,an<0当n≥8时,an>0|a1|+|a2|
/>n≥2时,an=Sn/n+2(n-1)Sn=nan-2n(n-1)S(n-1)=(n-1)an-2(n-1)(n-2)Sn-S(n-1)=an=nan-2n(n-1)-(n-1)an+2(n-1)
a[n]等差数列所以:S[5]=5*a[3]=30,a[3]=6a[6]-a[3]=3d,2-6=3d,d=-4/3a[4]=6-4/3=14/3a[5]=6-2*4/3=10/3所以S[8]=4*(
若S10>0,则S10=(a1+a10)*10/2>0则2a1+9d>0.则d>-2a1/9同理S11
d≠0,|a11|=|a51|-a11=a51-(a1+10d)=a1+50da1=-30da20=a1+19d-30d+19d=22d=-2a1=60an=a1+(n-1)d=60+(n-1)*(-
兄弟,这道题肯定错了!而且错的地方是‘S3=Sn’,应该改为“‘S3=Sn’n为一个确切的数字”如果改为S3=S5;则:a4+a5=0即2a1+7d=0;由于a1=13,可得d=-26/7.这样就可以
an,Sn,an^2成等差数列2Sn=an^2+an2Sn-1=an-1^2+an-1相减2an=an^2+an-(an-1^2+an-1)(an+an-1)=(an-an-1)(an+an-1)若{
24=S4=a1+a2+a3+a4=2(a2+a3)=>a2+a3=12a2*a3=35=>a2=5,a3=7=>a1=3=>an=3+(n-1)*2=2n+1bn=1/an*a(n+1)=1/((2