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2009-09-26T16:42:29+02:00
A + b + c = 168


[ciąg geometryczny]
b/a = c/b

[ciąg arytmetyczny]
a = a₁
b = a₅ = a₁ + 4r
c = a₂₁ = a₁ + 20r

a + b + c = 168 => a₁ + a₁ + 4r + a₁ + 20r = 168
3a₁ + 24r = 168 /:3

a₁ + 8r = 56

[wracamy do ciągu geometrycznego]
b/a = c/b
(a₁ + 4r)/a₁ = (a₁ + 20r)/(a₁ + 4r)
(a₁ + 4r)² = a₁(a₁ + 20r)
a₁² + 8a₁r + 16r² = a₁² + 20a₁r
16r² - 12a₁r = 0 /:4
4r² - 3a₁r = 0
4r(r - 3a₁/4) = 0
I. r = 0 ∨ II. r = a₁/12

2 przypadki:

I. r = 0 ∧ a₁ + 8r = 56 => a₁ = 56

a = a₁
b = a₅ = a₁ + 4r
c = a₂₁ = a₁ + 20r

a = 56
b = 56
c = 56
[ciąg geometryczny, gdzie q = 1 oraz ciąg arytmetyczny o r = 0]

v II. r = 3a₁/4 ∧ a₁ + 8r = 56 => a₁ + 6a₁ = 56 => a₁ = 8

a = a₁ = 8
b = a₅ = a₁ + 4r = a₁ + 3a₁ = 32
c = a₂₁ = a₁ + 20r = a₁ + 15a₁ = 128

a = 8
b = 32
c = 128
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