等比数列a3=5 2,且S3=15 2,则Q
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a(n)=a+(n-1)d,d不为0.[a(3)]^2=a(1)a(9)=a[a+8d]=[a+2d]^2,0=a^2+8ad-a^2-4ad-4d^2=4ad-4d^2=4d(a-d),0=a-d.
这是因为立方差公式推导的结果(a-b)(a²+ab+b²)=a³-b³这个知识点应该是初中二年级就学到的知识.
a3=2S2+1a4=2S3+1两式相减,得a4-a3=2(S3-S2)=2a3即,a4=3a3则,q=a4/a3=3所以,公比是3再问:S3-S2=a3?再答:对S3=a1+a2+a3S2=a1+a
a4-a3=2s3+1-(2s2+1)=2s3-2s2=2a3a4=2a3+a3=3a3q=3
依据题意,有2*3a2=a1+3+a3+4=7+a1+a3=7+a1+a2+a3-a2=7+7-a2=14-a2.2*3a2=14-a26a2=14-a27a2=14.a2=2.s3=a1+a2+a3
(1)∵a1,a3,a7成等比数列.∴a32=a1a7,即(a1+2d)2=a1(a1+6d),化简得d=12a1,d=0(舍去).∴S3=3a1+3×22×12a1=92a1=9,得a1=2,d=1
1)因为an为等差,故a1+a3=2*a2;所以S3=a1+a2+a3=a2+(a1+a3)=3*a2=12;故:a2=4;设等差数列{an}的等差为d(d>0);则a1=a2-d=4-d;a3=a2
S3=a1+a2+a3=3a2=12a2=4设公差为d,则a1=a2-d=4-da3=a2+d=4+d2a1、a2、a3+1成等比数列,则a2²=(2a1)(a3+1)2(4-d)(4+d+
等差6a2=a1+3+a3+4=a1+a3+7a1+a3=6a2-7S3=a1+a2+a3=7所以6a2-7+a2=7a2=2所以a1+a3=5即a2/q+a2q=5所以两边乘q2+2q²=
设公比为q,则a3=a1q^2a5=a1q^4由题意得2(S5+a5)=S3+a3+S4+a4即2(a1+a2+a3+a4+a5+a5)=a1+a2+a3+a3+a1+a2+a3+a4+a4整理得4a
S3应该是等于12吧{an}是等差数列S3=a1+a2+a3=3a2=12a2=4设公差为da1=4-da3=4+d2a1,a2,a3+1成等比数列(a2)^2=2a1·(a3+1)4^2=2(4-d
1)因为an为等差,故a1+a3=2*a2;所以S3=a1+a2+a3=a2+(a1+a3)=3*a2=12;故:a2=4;设等差数列{an}的等差为d(d>0);则a1=a2-d=4-d;a3=a2
S3=a1+a2+a3=a1+(a1+d)+(a1+2d)=3a1+3d=12所以a1+d=4(1)2a1,a2,a3+1成等比数列则a2²=2a1*(a3+1)即(a1+d)²=
.(1)S3=A1+A2+A3=72*3A2=(A1+3)+(A3+4)由以上2个方程解得:A2=2A1=2/qA3=2*q所以2/q+2+2*q=7解得:q=2An=2^(N-1)a(3n+1)=2
s3=a1+a2+a3=a1+a1q+a1q^2=1+q+q^21+q+q^2=7q^2+q-6=0(q-2)(q+3)=0q=2或q=-3当q=2时a1/a2+a2/a3+a3/a4+a4/a5+a
设等差数列{an}的公差为d,由题意得a22=2a1(a3+1)3a1+3×22d=12,解得a1=1d=3或a1=8d=−4,∴sn=12n(3n-1)或sn=2n(5-n).
(Ⅰ)设数列{an}的公比为q,显然q≠1,由题意得a1(1-q4)1-q+a1(1-q3)1-q=2a1(1-q2)1-qa3q+a3+qa3=-18,解得q=-2,a3=12,故数列{an}的通项
a1*a3=a2^2=4a2=2a1*q=2(1)s3=a1(1-q^3)/1-q=7(2)(1)÷(2)得2+2q+2q^2=7q得q=1/2舍q=2a1=1a5=a1*q^4=1*2^4=16
根据题意可列出方程组:a1+a2+a3=76a2=a1+3+a3+4a2×a2=a1×a3解得a1=1,a2=2,a3=4或者a1=4,a2=2,a1=1因为公比大于1,所以舍弃第二组答案,结果为a1