Question

proszę o rozwiązanie trzech zadań z ciągów geometrycznych, daję naaj! :)

Answer 1

z.1

a1 = 1

a2 = 1/2

q = 1/2

S10 = a1*[1 - q^10]/[1 - q] = 1*[1 - (1/2)^10]/[ 1 -1/2]

S10 = [ 1 - 1/(2^10)]/(1/2) = 2* [ 1 - 1/1024] = 2*[ 1024/1024 - 1/1024]

S10 = 2*(1023/1024) = 2046/1024 < 2

Odp.Nie

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z.2

a1 = 27

a3 = 12

a) a3 = a1*q^2 --> q^2 = a3: a1 = 12/27 = 4/9

zatem  q = 2/3

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b) an = a1*q^(n-1) = 27*(2/3)^(n-1) = 27*(2/3)^n * (2/3)^(-1)

an = 27*(3/2)*(2/3)^n = (81/2)* (2/3)^n

Odp. an = (81/2)*(2/3)^n

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c) S7 = a1*[1 - q^7]/[1 - q] = 27*[1 - (2/3)^7]/[1 - 2/3]

S7 = 27*[1 - 128/2187]/(1/3) = 27*3*(2059/2187) = 2059/27

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z.3

a1 = 8

q = 1,5 = 3/2

Sn = 65

Sn = 8*[1 - (3/2)^n]/[ 1 - (3/2)] = 8*[1 - (3/2)^n] /(-1/2)

Sn = -16*[ 1 -(3/2)^n]

czyli

-16 *[ 1 - (3/2)^n] = 65

-16 + 16*(3/2)^n = 65

16*(3/2)^n = 81

(3/2)^n = 81/16

n = 4

Odp. Trzeba dodać 4 wyrazy tego ciągu.

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