Question

w ciągu geometrycznym An dane są: a₁=32 i a₄= -4

a) Oblicz iloraz tego ciągu

b) oblicz szósty wyraz tego ciągu

c) oblicz sumę sześciu wyrazów tego ciągu

Answer 1

a1 = 32

a4 = - 4

zatem a4 = a1*q^3

a)q^3 = a4 : a1 = (-4) : 32 = - 1/8

q = - 1/2

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b) a6 = a1*q^5 = 32*( - 1/2)^5 = 32*(- 1/32) = - 1

a6 = - 1

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c)

S6 = a1 *[ 1 - q^6]/[ 1 - q]

S6 = 32*[ 1 - (-1/2)^6]/[ 1 - ( -1/2)]

S6 = 32*[ 1 + 1/64]/[1 + 1/2]

S6 = 32*[ 65/64]/[ 3/2]

S6 = 32*[65/64]*[ 2/3]

S6 = 65/3 = 21  2/3

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Answer 2

a) a4=a1*q³

-4=32*q³

q³=-4:32

q³=-0.125 /∛

q=-0.5

 

b) a6=a1*q^5

a6=32*(-0.5)^5

a6=32*(-0.03125)

a6=-1

 

c) s6=a1* [(1-q^n):(1-q)]

s6=32*[(1-(-0.5^6)):(1-(-0.5))]

s6=32*[(1-0.015625):(1.5)

s6=32*(0.984375:1.5)

s6=32*0.65625

s6=21