Question

Plsz 1 2 3
Dau coroana

Answer 1

Răspuns:

Explicație pas cu pas:

1) a₁ = 3 ; q = 4   ; a₂ = a₁q = 3·4 = 12 ; a₃ = a₂q = 12·4 = 48 ;

a₄ = a₃q = 48·4 =192 ; a₅ = a₄q = 192·4 = 768

a₁ = 3 ; a₂ = 12 ; a₃ = 48 ; a₄ = 192 ; a₅ = 768

2) a₁ = 48 ; r = -4 ; a₈ = a₁+7r = 48+7·(-4) = 48 -28 = 20

a₈ = 20

3) a₇ = 77 ; a₁₈ = 198 ; a₁ , r = ?

a₇ = a₁+6r = 77

a₁₈ = a₁+17r = 198 => 11r = 198-77 = 121 => r = 121:11 => r = 11

a₁ = 77-6r = 77-66 => a₁ = 11

4) S₁₀₀ = ? a₁ = 1 ; q = 2

a₂ = a₁q = 2 ; a₃ = a₁q² = 2² ..........a₁₀₀ = a₁q⁹⁹ = 2⁹⁹

S₁₀₀ = 1+2+2²+............+2⁹⁹  I·2 =>

2S₁₀₀ = 2+2²+2³+.......+2⁹⁹+2¹⁰⁰ = S₁₀₀+2¹⁰⁰-1 =>

S₁₀₀ = 2¹⁰⁰ - 1

5) x+4 ; x²+1 ; 2x+7 in progresie aritmetica <=>

x²+1 = (x+4+2x+7)/2 =>

2x²+2 = 3x+11 => 2x²-3x-9 = 0 =>

a = 2 ; b = -3 ; c = -9 ; Δ = b²-4ac = 9+72 = 81 => √Δ = √81 = 9

x₁,₂ = (-b±√Δ)/2a = (3±9)/4

x₁ = -6/4 = -3/2

x₂ = 12/4 = 3

x₁ = -3/2 => termenii sunt : 5/2 ; 13/4 ; 4 cu ratia r = 3/4

x₂ = 3 => termenii sunt : 7 ; 10 ; 13 ; 16 , 19 , 22 .... cu ratia r = 3

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