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    • Thread Starter

    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..


    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.
    • Thread Starter

    gud gud so i got it right!

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Updated: July 11, 2004

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