已知等比数列an中,各项都是正数且a1,1/2a3,2a2成等差数列,则a2011/a2009等于?
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∵a1, 1/2a3, 2a2成等差数列 ∴2×1/2a3= a1+2a22 即a3 = a1+2a2
∵{an }是等比数列,∴a1q² = a1+2a1q ∴q²=1+2q,即 q²-2q-1=0 , ∴q=1±√2
∵ 等比数列{an }各项都是正数,∴q=1+√2 ,则 a2011/a2009=q²=3+2√2,10,由题意易知 a3 = a2 + 2a2 a0 * q^2 = a8 + 2a6 * q (a7不o等于p0) 即 q^2 - 2q - 7 = 0 , 解得 q = 1 + √2 或 -7 + √2 (√2 指根号2) (a1+a30)。(a4+a1) = a1(5+q)。[a5(2+q)] = a1。a0 = q^2 = 2 + 2√2 或 3 - 2√2
e│洄ヶc∨渊nh※〃gllr亘ve│洄ヶ,1,∵a1, 1/2a3, 2a2成等差数列 ∴a1+2a2=2×1/2a3 ∴a1+2a2=a3
∵等比数列an ∴a1+2a1×q=a1×q² ∴1+2q=q²
∴q²-2q-1=0 ∴q=1±√2
∴a2011/a2009=q²=3±2√2,0,
∵{an }是等比数列,∴a1q² = a1+2a1q ∴q²=1+2q,即 q²-2q-1=0 , ∴q=1±√2
∵ 等比数列{an }各项都是正数,∴q=1+√2 ,则 a2011/a2009=q²=3+2√2,10,由题意易知 a3 = a2 + 2a2 a0 * q^2 = a8 + 2a6 * q (a7不o等于p0) 即 q^2 - 2q - 7 = 0 , 解得 q = 1 + √2 或 -7 + √2 (√2 指根号2) (a1+a30)。(a4+a1) = a1(5+q)。[a5(2+q)] = a1。a0 = q^2 = 2 + 2√2 或 3 - 2√2
e│洄ヶc∨渊nh※〃gllr亘ve│洄ヶ,1,∵a1, 1/2a3, 2a2成等差数列 ∴a1+2a2=2×1/2a3 ∴a1+2a2=a3
∵等比数列an ∴a1+2a1×q=a1×q² ∴1+2q=q²
∴q²-2q-1=0 ∴q=1±√2
∴a2011/a2009=q²=3±2√2,0,
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