在等比数列An中,公比q属于(0,1),且a1a5 2a3a5 a2 a8
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![在等比数列An中,公比q属于(0,1),且a1a5 2a3a5 a2 a8](/uploads/image/f/3263188-4-8.jpg?t=%E5%9C%A8%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97An%E4%B8%AD%2C%E5%85%AC%E6%AF%94q%E5%B1%9E%E4%BA%8E%280%2C1%29%2C%E4%B8%94a1a5+2a3a5+a2+a8)
我猜你的题目给出的条件是a(n+2)=a(n+1)+2an,就像楼上所列正解如下a3=a2+2a1=2a1+1a4=a3+2a2=2a1+1+2=2a1+3又an为等比数列,a2=a1*q,a3=a1
S4=a1(1-q^4)/(1-q)=5a1(1-q^2)/(1-q)1+q^2=5q^2=4因为q
{an}为等比,各项均为正数,则:q>0a5=a3q²,a6=a3q³a3,a5,a6成等差数列则:2a5=a3+a6即:2a3q²=a3+a3q³约去a3得:
因为A1,a2,a7成等比数列,又{an}是等差数列有(a1+d)^2=a1*(a1+6d)解得d=4a1q=a2/a1=(a1+d)/a1=5a1/a1=5所以公比为5
设A1A2=a则:由于在数列{An}中An小于0故a>0,且An+1An+2/AnAn+1>0即q>0;由题中:2AnAn+1+An+1An+2>An+2An+3得2aq^(n-1)+aq^n>aq^
由a4a1=q3=648=8可得q=2.
设等差数列{an}的公差为d,则可得(a1+d)2=a1(a1+3d)解得a1=d或d=0∴公比q=a2a1=2或1.故答案为:2或1.
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这个图片不知道行不行啊再问:{an+1}为等比数列怎麽会有An+1+An-1=An再答:这是按照上面的公式算出来的啊,是等于2An因为an是等比数列,所以an+1*an-1=an*an
2a3=a2+a52a₁q²=a₁q+a₁q⁴q⁴-2q²+q=0q(q-1)(q²+q-1)=0q≠0,q≠
a1+an=66a2an-1=a1an=128所以a1=2,an=64或a1=64,an=2(舍去)an=a1q^(n-1)=64q^(n-1)=32Sn=a1(1-q^n)/(1-q)=126,即2
a1(1+q)=1,a1q^2(1+q)=4q^2=4,q=-2a4+a5=a1q^3(1+q)=(a3+a4)*q=-8
a4*a(n-3)=128,由等比数列的性质,可知:a1*an=128还知道a1+an=66,所以,可以二元一次方程解情况1:a1=2,an=64;或情况2:a1=64,an=2先做第一种情况,之后,
a1a2a3a4a5=(a3)^5=q^10=a11,因此m=11
1.(a5)^2=a3a7=1/81因为a1=9>0,q0a5=1/92.s4=a1(1-q^4)/(1-q)=4s8=a1(1-q^8)/(1-q)=16s8/s4=(1-q^8)/(1-q^4)=
在等比数列{an}中,由a5=a2q3,又a2=8,a5=64,所以,q3=a5a2=648=8,所以,q=2.故选A.
s3:s2=(a1+a2+a3)/(a1+a2)=1+1/(1/q^2+1/q)=3/2所以1/q^2+1/q=21/q=-2或1q=-1/2或1再问:这是一道选择题,题中没有这个答案呀再答:有神马选
A1*q^9=1,A1*q^18=512,两式相除可得q^9=512,可解得q值.
∵等比数列{an}中,公比q=12,且log2a1+log2a2+…+log2a10=55=log2(a1a2…a10)=log2 (a1a10) 5,∴(a1a10)5=255,
等比数列an中a1=1/2,a4=4则公比q=(a4/a1)开3次方=8开3次方=2a1+a2+…+an=Sn=a1(1-q^n)/(1-q)=1/2(1-2^n)/(1-2)=2^(n-1)-1/2