1.Dany jest rosnący ciąg geometryczny, w którym a₁ = 12, a₃= 27. Oblicz iloraz tego ciągu oraz oblicz wyraz a₆
2 zadanie jest w załączniku
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z.1
a1 = 12, a3 = 27
a3 = a1*q^2
a3/a1 = [a1*q^2]/a1 = q^2
ale a3/a1 = 27/12 = 9/4
zatem q^2 = 9/4
q = 3/2
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a6 = a1*q^5 = 12* (3/2)^5 = 12*( 243/32) = 729/8 = 91 1/8
a6 = 91 1/8
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z.2
Sn = n *( 2n + 1 )
S1 = a1 = 1*(2*1 +1) = 1*3 = 3
S2 = a1 + a2 = 2*(2*2 +1) = 2*5 = 10
zatem a2 = 10 - a1 = 10 - 3 = 7
S3 = a1 + a2 + a3 = 3*(2*3 + 1) = 3*7 = 21
a3 = 21 - a1 - a2 = 21 - 10 = 11
a3 - a2 = 11 - 7 = 4 oraz a2 - a1 = 7 - 3 = 4
Mamy ciąg arytmetyczny o różnicy r = 4 oraz a1 = 3
zatem a6 = a1 + 5*r = 3 + 5*4 = 23
Zatem
S6 = 0,5 *[a1 + a6] *6 = 3*[ 3 + 23] = 3*26 = 78
lub S6 = 6*[2*6 + 1] = 6*13 = 78
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