等差数列AN各项均为正数公差D为正整数A1=3
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由于为等比数列,只要连续3项就可确定数列的首项和公比!故只需要讨论4项删去某一项后剩3项即可!故只要讨论a1,a2,a3,a4即可!(1)删掉首项:a2,a3,a4a3^2=(a3-d)(a3+d)d
{根号Sn}的公差为d的等差数列√Sn=√S1+(n-1)dSn=[√S1+(n-1)d]^2=S1+(n-1)^2d^2+2√S1(n-1)dS2=S1+d^2+2√S1dS3=S1+4d^2+4√
√S2=√S1+d=√a1+d,S2=(√a1+d)²=a1+d²+2d√a1a2=S2-S1=a1+d²+2d√a1-a1=d²+2d√a1a3=2a2-a1
这是今年江苏卷上的题目…………(1)设根号Sn=d*n+HSn=d^2*n^2+2*d*H*n+H^2a1=S1=d^2+2*d*H+H^2a2=S2-S1=3*d^2+2*d*Ha3=S3-S2=5
1、√S1=√a1√S2=√(a1+a2)=√a1+2(1)√S3=√(a1+a2+a3)=√(3a2)=√a1+4(2)由(1)得a1+a2=a1+4√a1+4√a1=(a2-4)/4代入(2)√(
√Sn=√S1+(n-1)d√S2=√S1+d√S3=√S1+2d第2个式子两边平方a1+a2=a1+2(√a1)d+d^2第3个式子两边平方a1+a2+a3=a1+4(√a1)d+4d^2两个式子相
s1=a1;s2=a1+a2;s3=a1+a2+a3=3a2根号s3=根号s1+2d=根号s2+d化简得a2=3a1代入等差数列可求得d=根号a1sn=(nd)²an=sn-sn-1=(2n
结果是an=4(2n+1);首先由s1,s2,s3的关系可列出两个方程,关于a1,a2,a3.和已知的2a2=a1+a3联立,求出a1=4.接下来,利用根号sn是等差数列,推导出s(n)和a1的关系,
(1)根据题意,设公差为d则a3=a1+2d=2d+1a9=a1+8d=8d+1有(2d+1)^2=8d+1d=1故通项:an=n(2)根据题意,设公比为q则b2=qb3=q^2有q-0.5q^2=0
设等差数列的公差为d,则a3=a5-2d=6-2d,an1=a5+(n1-5)d=6+(n1-5)d.∵a3,a5,an1成等比数列,∴a52=a3an1化简即(6n1-42)d-2(n1-5)d2=
d=50/(n-1),注意,题目中说个各项均为正整数,所以d也只能是正整数,因此,d只能是1,2,5,10,25,50这些数,此时n分别为51,26,11,6,3,2n+d最小就是16
1.若n=4时,则原数列为a1,a2,a3,a4.⑴若删去a1,则a3∧2=a2×a4,→d=0,矛盾⑵若删去a2,→a5=0矛盾⑶若删去a3→a1=d→a1/d=1⑷若删去a4→d=0矛盾综上所述,
∵等差数列{an}各项均为正数,且a2+a3+a4+a5=34,a2a5=52,∴a2+a5=17,a2a5=52,解得a2=4,a5=13.∵a5=a2+3d,∴13=4+3d,解得d=3.故选:C
可用递推法:2Sn=An+An*An递推2Sn-1=An-1+An-1*An-1两市相减,得:An+An-1=An*An-An-1*An-1因为An为正数,所以An-An-1=1之后求An,然后用求和
∵{log2an}是公差为-1的等差数列∴log2an=log2a1-n+1∴an=2log2a1−n+1=a1•2−n+1∴S6=a1(1+12+…+132)=a1•1−1261−12=38,∴a1
{an}各项为整数,所以d为整数,且d≠0a5=6→a2=6-3da10=6+5da2*a10=(6-3d)(6+5d)>10→36+2d-15d^2>10→15d^2-2d-260d=-1时,a2=
再问:为什么2lga2=lga1+lga4a2²=a1a4再答:这是对数函数的运算规则啊!你没学过对数函数?
由a1=1,得到an=a1+(n-1)d=1+(n-1)d=51,即(n-1)d=50,解得:d=50n−1,因为等差数列的各项均为正整数,所以公差d也为正整数,因此d只能是1,2,5,10,25,5
根号Sn的通项公式是nSn=n^2an=Sn-Sn-1=n^2-(n-1)^2=2n-1