www.1862.net > 已知各项为正数的等比数列{An}满足:A7=A6+2A5,若...

已知各项为正数的等比数列{An}满足:A7=A6+2A5,若...

∵{an}为等比数列 ∴an=a1*q^(n-1) ∵a7=a6+2a5 ∴q²=q+2 ∴q=2或者q=-1 ∵an>0 ∴q=2 要使得√(am*an)=2√2*a1 即√(a1*q^(m-1)*a1*q^(n-1))=2√2*a1 √(q^(m+n-2))=2√2=√(2^3) 2^(m+n-2)=2^3 m+n-2=3 m+n=5 ∴m/5+n/5=1 ∴1/m+4/n=(1/m+4/n)*(m/5+n/5)...

q=2 或1/2; 根据等比数列性质: a1a5=a2a4=(a3)*(a3);a4a6=a3a7=(a5)*(a5);a2a6=a3a5; 所以上式可变为两个方程组(a3+a5)*(a3+a5)=100 (a3-a5) *(a3-a5)=36 可解得 1.a3=8,a5=2; 2.a3=2,a5=8; (3.a3=-2,a5=-8; 4.a3=-8;a5=-2;不合题意舍去) 可...

∵等比数列{an}的各项均为正数,a6a4+2a8a5+a9a7=36,∴a52+2a8a5 +a82=25,即(a5+a8)2=36,∴a5+a8=6,故选C.

由等比数列得性质可得a1a10=a2a9=…=a5a6,又∵a5a6+a4a7=8,∴a1a10=a2a9=…=a5a6=4,∴log2a1+log2a2+…+log2a10=log2(a1?a2?…a10)=log2(a1a10)5=log245=log2210=10故答案为:10

a2+a3=a1(q+q2)=12则q+q2=6解之得q=-3或2,又各项均为正数所以q=2,所以a4+a5+a6=(a1+a2+a3)×q的立方=14×8=112

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