Dany jest ciąg arytmetyczny an taki ,że :
a1=3 oraz a3=7 .
Korzystamy z wzoru :
an=a1+(n-1)r
a3=a1+2r
7=3+2r
2r=7-3
2r=4|:2
r=2
a20=a1+19r
a20=3+19·2
a20=41
Sn=[(a1+an)/2] ·n
S14=(a1+a14)/2· 14
S14=(a1+a14)·7 , gdzie a14=a1+13r czyli a14=3+13·2=29
S14=(3+29)·7
S14=224
2. ( 2 , x , y , 54 ) - ciąg geometryczny
a1=2
a4=54
Korzystamy z wzoru :
an=a1q^(n-1)
a4=a1q³
54=2q³ |:2
q³=27
q=∛27
q=3
x=2q
x=2·3=6
y=2q²
y=2·3²
y=18
Odp. x=6 , y=18 .
Odpowiedź:
zad 1
a₁ = 3
a₃ = a₁ + 2r = 7
a₁ + 2r = 7
3 + 2r = 7
2r = 7 - 3 = 4
r = 4 : 2 = 2
a₁₄ = a₁ + 13r = 3 + 13 * 2 = 3 + 26 = 29
a₂₀ = a₁ + 19r = 3 + 19 * 2 = 3 + 38 = 41
S₁₄ = (a₁ + a₁₄) * 14/2 = (3 + 29) * 7 = 32 * 7 = 224
zad 2
a₁ = 2
a₂ = x
a₃ = y
a₄ = 54
a₄/a₃ = a₃/a₂ = a₂/a₁
54/y = y/x = x/2 dla x ≠ 0 ∧ y ≠ 0
y² = 54x
x² = 2y
y = x²/2
y² = 54x
(x²/2)² = 54x
x⁴/4 = 54x | * 4
x⁴ = 4 * 54x = 216x
x⁴ - 216x = 0
x(x³ - 216) = 0
x = 0 ∨ x³ - 216 = 0
x ≠ 0 więc:
x³ - 216 = 0
x³ = 216
x = ∛216 = 6
y = x²/2 = 6²/2 = 36/2 = 18
a₁ = 2
a₂ = x = 6
a₃ = y = 18
a₄ = 54
sprawdzenie
a₄/a₃ = a₃/a₂ = a₂/a₁
54/18 = 18/6 = 6/2
3 = 3 = 3