Răspuns:
Explicație pas cu pas:
a₁ , a₂ , a₃ , a₄ , a₅ , a₆
a₃² = a₆
a₁+a₂ = 3a₃/4 <=>
a₁+a₁q = 3a₁q²/4 =>
4a₁(1+q) = 3a₁q² => q² = 4(q+1)/3
(a₁q²)² = a₁q⁵ <=> a₁²q⁴ = a₁q⁵ => a₁ = q
q ; q² ; q³ ; q⁴ ; q⁵ ; q⁶
3q² = 4q + 4 => 3q²-4q-4 = 0 ; a = 3 ; b = -4 ; c = -4
Δ = b²-4ac = (-4)²-4·3·(-4) = 16+48 = 64 =>
√Δ = √64 = 8
q₁,₂ = (-b±√Δ)/2a = (4±8)/6
q₁ = -4/6 = -2/3
q₂ = 12/6 = 2
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progresia₁ : -2/3 ; 4/9 ; -8/27 ; 16/81 ; -32/243 ; 64/729
progresia₂ : 2 ; 4 ; 8 ; 16 ; 32 ; 64
