解答

✨ 最佳解答 ✨

29.
令三數為a ar a*r^2 (r > 0)
a + ar + a*r^2 = a (r^2 + r + 1) = 52
1/a + 1/(ar) + 1/(a*r^2)
= (r^2 + r + 1)/(a*r^2)
= (52/a)/(a*r^2)
= 52/(ar)^2
= 13/36
= 52/36*4
ar = 12

12/r + 12r = 52-12 = 40
3*r^2 - 10r + 3 = 0
(r-3)(3r-1) = 0
r = 3 or 1/3

r = 3,三數為4 12 36
r = 1/3,三數為36 12 4
三數中的最大數為36

30.
a1 = 1/5
a2 = 2/5
a3 = (5/2)(2/5)(1-2/5) = 3/5
a4 = (5/2)(3/5)(1-3/5) = 3/5
循環了 所以後面都是3/5
a101 - a100 = 0

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