Question

Rozwiązać to co w załączniku ;] Dużo pkt!

Answer 1

Załączniki:
z.558
a)
a1 =5 oraz a2 = 1
q = a2 : a1 = 1 : 5 = 1/5
a3 = a2*q = 1*(1/5) = 1/5
a4 = a3*q = (1/5)*(1/5) = 1/25
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b) a1 = -0,2 oraz a2 = -0,4
q = -0,4 : (-0,2) = 2
a3 = a2*q = -0,4*2 = -0,8
a4 = a3*q = -0,8*2 = -1,6
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c) a1 = 2/3 oraz a2 = -4/3
q = (-4/3) : (2/3) = (- 4/3)* (3/2) = -2
a3 = a2*q = (-4/3)*(-2) = 8/3
a4 = a3*q = (8/3)*(-2) = -16/3
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d) a1 = -5 oraz a2 = 5
q = 5 : (-5) = -1
a3 = 5*(-1) = -5
a4 = (-5) *(-1) = 5
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z.559
a) 1/4; 1/10; a3,a4
q = (1/10) : (1/4) =(1/10)*4 = 4/10 = 2/5
a3 = (1/10)*(2/5) = 2/50 = 1/25
a4 = (1/25)*(2/5) = 2/125
b)
a1, -3/2, 3/4,a4
q = (3/4):(-3/2) = (3/4)*(-2/3) = - 1/2
a1 = a2: q = (-3/2) :(-1/2) = (3/2)*2 = 3
a4 = a3*q = (3/4)*(-1/2) = -3/8
c)
a1,2,a3,4
a3 =√(2*4) = 2√2
q = a3 :a2 = 2√2 : 2 = √2
a1 = a2:q = 2 : √2 = √2
d)
-1, a2, -1/9,a4
a2 =√(-1*(-1/9) = √(1/9) = 1/3
q = a2 : a1 = (1/3) : (-1) = -1/3
a4 = a3*q = (-1/9)*(-1/3) = 1/27
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z.560
a) a1 = - 16, q = 1/2
an = a1*q^(n -1) = -16*(1/2)^(n-1)
a5 = -16*(1/2)^(5-1) = -16*(1/2)^4 = -16*(1/16) = -1
b)
a1 = 1/32, q = -2
an = (1/32)*(-2)^(n-1)
a5 = (1/32)*(-2)^4 = (1/32)*16 = 1/2
c)
a1 = 12, q = 0,1
an = 12*(0,1)^(n -1)
a5 = 12*(0,1)^4 = 12*0,0001 = 0,0012

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z. 561
a)
3,5; 7; 14
a1 = 3,5 oraz q = 7 : 3,5 = 2
an = 3,5*2^(n -1)
b)
-20,10, -5
a1 = -20 oraz q = 10 : (-20) = -1/2
an =(-20)*(-1/2)^(n -1)
c)
1/10, 1/100, 1/1000
a1 = 1/10 oraz q = 1/10
an = (1/10)*(1/10)^(n -1) = (1/10)^n
d)
2, √2,1
a1 = 2 oraz q = 1/√2
an =2*(1/√2)^(n -1)
e)
-3,0,0
To nie jest ciąg geometryczny
f)
6,6,6
q = 1
an = 6
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z.562
a) 1/4, -1/2 , ... , x = -32
a1 = 1/4 oraz q = (-1/2) :(1/4) = (-1/2)*4 = - 2
an = -32
-32 = (1/4)*(-2)^(n -1) ----> (-2)^(n -1) = -32*4 = - 128
n -1 = 7 , bo (-2)⁷ = - 128
zatem n = 8
x = -32 jest 8 wyrazem tego ciagu
b)
9/5, 3/5, .. , x = 1/135
a1 = 9/5 oraz q = (3/5) : (9/5) = (3/5)*(5/9) =1/3
an = 1/135
1/135 = (9/5)*(1/3)^(n -1) --> (1/3)^(n -1) = (1/135)*(5/9)
(1/3)^(n -1) = 1/243
n-1 = 5 , bo (1/3)^5 = 1/243
zatem n = 6
x = 1/135 jest 6 wyrazem tego ciągu
c)
16,12, ... , x = 27/4
a1 = 12 oraz q = 12 /16 = 3/4
27/4 = 16*(3/4)^(n -1) ---> (3/4)^(n -1) =( 27/4) : 16 = 27/64
n -1 = 3, bo (3/4)³ = 27/64
zatem n = 4
x = 27/4 jest 4 wyrazem tego ciągu
d)
π/81, π²/27, ... , x = 81 π⁹
a1 = π/81 oraz q = 3 π
81 π⁹ = (π/81)*(3 π) ^(n -1) --> (3 π)^(n -1) = 81² π⁸
(3 π)^(n-1) = (3 π)⁸ --> n -1 = 8
zatem n = 9
x = 81 π⁹ jest 9 wyrazem tego ciagu
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