等比数列an中,a2,a8是方程x²-34x 64=0的两根

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

∵{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∴

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

a1+a2+a3=3a7+a8+a9=a1*q^6+a2*q^6+a3*q^6=(a1+a2+a3)q^6q^6=(a7+a8+a9)/(a1+a2+a3)=192/3=64q=2或q=-2若q=2a

等差则2×1/2a3=a1+2a2a3=a1+a2所以a1q²=a1+2a1q除以a1q²-2q-1=0都是正数则q>0所以q=1+√2a9/a7=q²a10/a8=q&

a(n)=a+(n-1)d,n=1,2,...[a(4)]^2=[a(3)+d]^2=(6+d)^2=a(2)*a(8)=[a(3)-d][a(3)+5d]=(6-d)(6+5d),36+12d+d^

240

a2=a1qa8=a1q^7a5=a1q^42a8=a2+a52a1q^7=a1q+a1q^42q^6=1+q^32q^6=1+q^32q^6-q^3-1=0(2q^3+1)(q^3-1)=0q^3=

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/q+qa2=5a2+a2q^2=10a2+a2q^2=5q5q=10q=2a1+a1q^2=65a1=5a1=1a8=1×2^7=128

a1,a2,a4成等比数列(a2)^2=a1*a4(a2)^2=(a2-d)(a2+2d)(a2)^2=(a2)^2+a2d-2d^2a2d=2d^2a2=2d(a1+a2+a4)／(a2+a4+a8

因为a1a5=20,a2+a4=-12｛a}等比,所以a2a4=20,a2,a4是方程：x^2+12x+20=0的根x=-2或x=-101.a2=-2,a4=-10q^2=a4/a2=5a8=a4*q

1.-14582.13.5

1.等比数列an中,a5恰好是a2与a8的等比中项,故有a5^2=a2*a8,故得a8=-432.2（看题目,设an的通项简单的多）令an=a1+nd,故有3a1+3d=30,（a1-5）(a1+2d

an=a1q^(n-1)a2=a1q=-2a5=a1q^4=54q^3=-27q=-3a1=2/3再问：是求a18不是a8

192a2=a1*q=3a5=a1*q^4=24所以q=2,a1=3/2a8=a1*q^7=192

log2（a1）+log2（a2）+……+log2（a8）=log2(a1×a2×…×a8)∵等比数列∴a1a8=a2a7=a3a6=a4a5=32∴log2(a1×a2×…×a8)=log2(32^

a2×a10=a1q×a1q^9=a1^2×q^10a4×a8=a1q^3×a1q^7=a1^2×q^10所以a4×a8=a2×a10=12

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