等比数列{an}中,已知a2=6,a5=16, , 则a8=( )

a2=a1q=18a4=a1q^3=8a4/a2=q^2=9/4q=3/2或q=-3/2a1=12或a1=-12

∵{an}是等差数列∴a2+a8=a3+a7=2a5∴3a5=9解得:a5=3∴a3+a7=6.(1)a3a7=-7.(2)联立(1)(2):(a3-7)(a3+1)=0a3=7或-1a7=-1或7∴

a(1)=2^1-1=1,2^n-1=a(1)+a(2)+...+a(n),2^(n+1)-1=a(1)+a(2)+...+a(n)+a(n+1)=2^n-1+a(n+1),a(n+1)=2^(n+1

设公比为q.a1+a2=a1(1+q)=9a1a2a3=a1^3q^^3=27a1q=3a1=3/q(3/q)(1+q)=93q+3=9q6q=3q=1/2a1=3/(1/2)=6Sn=a1(1-q^

log3a1+log3a2+...+log3a8=log3[a1a2...a8]=log3[(a1a8)(a2a7)..(a4a5)]=log3[9^4)=log3(3^8)=8

由等比数列性质得到a1a3=a2的平方a1a2a3=a2的立方=8a2=2a1+a2+a3=a2/q+a2+a2*q=-3q=-2q=-1（不满足舍去）当q=-2时候a4=a2*q的平方=2*4=8

a1+a2=20,a3+a4=q^2(a1+a2)=40q^2=2a5+a6=q^2(a3+a4)=2*40=80a1+a2=3,a2+a3=q(a1+a2)=6q=2a1+a1*2=3a1=1a7=

因为a1+a2+a3=7,a1a2a3=8又因为等比数列{an},那么a2*a2=a1a3,那么a1a2a3=a2a2a2=8,所以a2=2,那么a1+a3=5,同时a1a3=4所以a1=1,a3=4

a1=a1a2=a1qa3=a1q^2a1(1+q+q^2)=14a1a2a3=a1^3q^3=64a1q=4a1=4/q代入,4(1+q+q^2)=14q整理,得2q^2-5q+2=0(q-2)(2

（1）a1a2a3=(a2/q)a2(a2q)=(a2)^3=27a2=3（2）a1+a2+a3=13a2/q+a2+a2q=133/q+3+3q=13整理,得3q²-10q+3=0(3q-

因为a1+a2+a3=7,a1a2a3=8又因为等比数列{an},那么a2*a2=a1a3,那么a1a2a3=a2a2a2=8,所以a2=2,那么a1+a3=5,同时a1a3=4所以a1=1,a3=4

a2＞a3=1q=a3/a20＜q＜1a1=a3/q²=1/q²＞1(a1+a2…+an)-(1/a1+1/a2…+1/an)＞0将数据带入不等式所以n的最大值为4再问：从“将数据

1．设公比为q,∵a2*q*q=a4∴18*q*q=8→→q^2=4/9→→q=2/3或q=-2/3.①当q=2/3时a1*q=a2→→a1*2/3=18→→a1=27.Sn=a1*(1-q^n)/(

设公比为qa2>a3=10

等比性质,a1a5=a2a4=(a3)²=1,a1a3=(a2)²>1,所以T5=(a1－1/a1)＋(a2－1/a2)＋(a3－1/a3)＋(a4－1/a4)＋(a5－1/a5)

a2=8,a5=64,所以q=2,a1=4,然后根据公式算出s10a10=a1×q^(10-1)；=4*2^9；=2048S10=(a1-a10×q)/(1-q)=(4-2048*2)/(1-2）;=

an=32*（3/8开6次方的n-2次方）Tn=log（2^n*a1*a2...an）问题转化为求a1*a2*...*an的值S=32^n*（3/8的n(n-2)/6次)所以Tn=log(64^n*（

a2=a1qa1+a1q=2√2a1=2√2/(1+q)a1*a1q=28q/(1+q)^2=24q=q^2+2q+1q^2-2q+1=0(q-1)^2=1q=1a1=√2an=√2

1]a5/a2=27=q^3q=3,a1=a2/q=1=>an=3^(n-1)2]S6=3(a1+a2+..+a6)-6x2=3(3^6-1)/2-12=364x3-12=1080

∵等比数列{an}中，a2+a7=66，a3a6=128，∴a2+a7=66，a2a7=128，∴a2=2，a7=64或a2=64，a7=2，∴q=2或q=12，∴an=2n-1或an=21-n．