Odpowiedź:
zad 1
a₁ = - 9
a₄ = a₁q³ = - 8/3
a⁴/a₁ = a₁q³/a₁ = q³
q³ = (- 8/3) : (- 9) = 8/3 * 1/9 = 8/27
q = ∛(8/27) = 2/3
lub
q³ - 8/27 = 0
q³ - (2/3)³ = 0
Korzystamy ze wzoru a³ - b³ = (a - b)(a² - ab + b²)
q³ - (2/3)³ = (q - 2/3)(q² - 2/3q + 4/9)
(q - 2/3)(q² - 2/3q + 4/9) = 0
q - 2/3 = 0 ∨ q² - 2/3q + 4/9 = 0
q = 2/3
q² - 2/3q + 4/9 = 0
Δ = (- 2/3) - 4 * 1 * 4/9 = 4/9 - 16/9 = - 12/9 = - 4/3
Δ < 0 więc brak miejsc zerowych
q = 2/3
zad 2
an = - 3n + 9 dla n ≥ 1
- 3n + 9 ≥0
- 3n ≥- 9
3n ≤ 9
n ≤ 9/3
n ≤ 3
n = { 1 , 2 , 3 }
Są trzy wyrazy nieujemne
zad 3
a₁ = - 7
a₂ = a₁ + r = - 7 + r
- 2 = - 7 + r
r - różnica ciągu = - 2 + 7 = 5
a₂₀ = a₁ + 19r = - 7 + 19 * 5 = - 7 + 95 = 88
S₂₀ = (a₁ + a₂₀) * 20/2 = (- 7 + 88) * 10 = 81 * 10 = 810
