# help with this sequence!

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#1
hi can anyone help with this question:

a1, a2, a3,....., an,..... are positive intergers such that for all n>=1:

a(n+1) > an, and a(an) = 3n.

a) find a1.
b) find a2, a3,..., a9

i get

2, 3, 6, 7, 8, 9, 12, 15, 18

although there are no answers in the book so i dunno... seems to work in my head..

hiroki
0
16 years ago
#2
a1 must be 1, 2 or 3 because the sequence is increasing and contains a 3.

It can't be 1 because then a(a1) = a(1) = 3 would contradict a(1) = 1. It can't be 3 because then a(a1) = a(3) = 3 would imply that the sequence is not increasing. So a1 = 2.

That gives us, using a(an) = 3n four times,

2 3 6 _ _ 9 _ _ 18.

Because the sequence is increasing we can fill in the first two blanks:

2 3 6 7 8 9 _ _ 18.

Using a(an) = 3n another two times,

2 3 6 7 8 9 12 15 18.
0
#3
gud gud so i got it right!

hiroki
0
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