CIĄG GEOMETRYCZNY.
1. Znajdź m dla którego liczby m-7, m+3, 4m-3 stanowią wyrazy ciągu geometrycznego
2. Oblicz: 3-9+27-...+2187
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zad. 1
a₁ = m-7
a₂ = a₁q = (m-7)q = m+3
a₃ = a₂q = (m+3)q = 4m-3
/*(-1)
-10q = -3m+6
(3m-6)(m-7) = 10m +30
3m² - 21m - 6m + 12 = 0
Δ= 1369-144=1225
√Δ=35
m₁ =
m₂ =
zad. 2
a₁ = 3
3*q₁= -9 ⇒ q₁= -3
-9* q₂ = 27 ⇒ q₂ = -3
q₁ = q₂ = q
n=7, bo 3⁷ = 2187