等比数列中,已知a1=1,a8=128,则S9=()
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由题意易知a3=a1+2a2a1*q^2=a1+2a1*q(a1不等于0)即q^2-2q-1=0,解得q=1+√2或-1+√2(√2指根号2)(a9+a10)/(a7+a8)=a9(1+q)/[a7(
a1*a1*q²*a1*q^10=(a1*q^4)³=8所以a5=a1*q^4=2所以a2*a8=(a5)²=4
log3a1+log3a2+...+log3a8=log3[a1a2...a8]=log3[(a1a8)(a2a7)..(a4a5)]=log3[9^4)=log3(3^8)=8
A1+A8=A1(1+q^7),A4+A5=A1(q^3+q^4)所以用作差法比较:(A1+A8)-(A4+A5)=A1(q^7-q^4-q^3+1)=A1[q^4(q^3-1)-(q^3-1)]=A
由a1,1/2a3,2a2成等差数列则有a3=a1+2a2,设等比数列(an)公比为q则有a2*q=a2/q+2a2因为数列(an)中各项都是正数所以两边同除以a2得q=1/q+2解得q=1+根号2或
an=a1q^(n-1)a1,(1/2)a3,2a2成等差数列2a2+a1=a32a1q+a1=a1q^2q^2-2q-1=0q=1+√2(a9+a10)/(a7+a8)=a1q^8(1+q)/[a1
a3=a1+2a2q^2=1+2qq^2-2q-1=0,q=1±√2因为等比数列{am}中,各项都是正数,所以公比为正数,即q=1+√2(a9+a10)/(a7+a8)=a1*q^8(1+q)/[a1
由题意易知a3=a1+2a2a1*q^2=a1+2a1*q(a1不等于0)即q^2-2q-1=0,解得q=1+√2或-1+√2(√2指根号2)(a9+a10)/(a7+a8)=a9(1+q)/[a7(
q^7=a8/a1=81÷1/27=3^7q=3所以S8=a1*)1-3^8)/(1-3)=3280/27再问:你们答案都不一样到底哪个是对的啊?
180a2+a5+a8=(a1+a4+a7)*q所以q=2a3+a6+a9=(a2+a5+a8)*q
(1)由a2=b2a8=b3a1=b1=1得1+d=q1+7d=q2(3分)∴(1+d)2=1+7d,即,d2=5d,又∵d≠0,∴d=5,从而q=6(6分)(2)∵an=a1+(n-1)d=5n-4
a4=a1q^3a5=a2q^3a6=a3q^3所以a4+a5+a6=q^3(a1+a2+a3)q^3=168/21=8同理a7=q4*q^3,……所以a7+a8+a9=q^3*(a4+a5+a6)=
a2/q+qa2=5a2+a2q^2=10a2+a2q^2=5q5q=10q=2a1+a1q^2=65a1=5a1=1a8=1×2^7=128
a1,1/2a3,2a2成等差数列2*1/2a3=a1+2a2a3=a1+2a2所以a1q²=a1+a1q两边除以a1q²=1+qq²-q-1=0(an)中各项都是正数,
等比数列{an},a1a2……a8=16,所以a1a8=a2a7=a3a6=a4a5=2原式=(1/a1+1/a8)+(1/a2+1/a7)+...+(1/a4+1/a5)=(a1+a8)/2+(a2
1、已知等比数列{an}中,a5=1/16,a8=-1/128,求q和a1,并写出它的通项公式.等比数列的通项公式为an=a1×q的(n-1)次方所以a5=a1×q的4次方=1/16,a8=a1×q的
题目:等比数列an中,a1,1/2a3,2a2等差数列,(a8+a9)/(a6+a7)=?a1、a3/2、2a2成等差数列,则a3=a1+2a2a1q²=a1+2a1q(a1不等于0)整理,
∵{an}为等比数列,∴an=a1*q^(n-1)设bn=1/an,则bn=1/a1×q^(1-n)∴b(n+1)/bn=q^[1-(n+1)]/q^(1-n)=q^(-1)∴{bn}为等比数列前8项