Jeśli 1/9, a,b,c,9 to pięć kolejnych wyrazów ciągu geometrycznego, to
A. a= -1/3, b=1 c=3
B. a=-1/9 b=1 c=3 lub a=1/9 b=1 c=-3
C. a=1/9, b=1, c=-3
D. a= 1/3, b=1, c=3 lub a= -1/3, b=1, c=-3
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a1 = 1/9
a2 = a = a1*q
a3 = b = a1*q^2
a4 = c = a1*q^3
a5 = a1*q^4 = 9
zatem
(1/9)*q^4 = 9 / * 9
q^4 = 81
q = - 3 lub q = 3
czyli
a = a1*q = (1/9)*(-3) = -1/3 lub a = (1/9)*3 = 1/3
b = a1*q^2 = (1/9)*9 = 1
c = a1*q^3 = (1/9)*(-3)^3 = - 3 lub c = (1/9)*3^3 = 3
Odp. D
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a₅=a₁q⁴
9=¹/₉q⁴
q⁴=9;¹/₉
q⁴=81
q=⁴√81
q=3 lub q=-3
dla q=3;
a=¹/₉×3=⅓
b=⅓×3=1
c=1×3=3
dla q=-3;
a=¹/₉×(-3)=-⅓
b=(-⅓)×(-3)=1
c=1×(-3)=-3
odp. d]