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Geometric Sequence Problem

  1. Jan 14, 2004 #1

    JDK

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    Hi,

    I have a relatively simple question. In this particular problem in my Math 30 Pure text book...

    10. Here are three levels in a school telephoning tree.

    Teacher
    Student 1 Student 2
    Student 3 Student 4 Student 5 Student 6

    a)At what level are 64 students contacted?

    ...I really need to know if there is a formula I can use to get the answer. I've tried using tn = (a)(r^(n-1)), and Sn = a((r^n)-1)/r-1 while replacing both tn and Sn with 64 and solving. Either way I end up with incorrect answers. The correct response, as said by the book is 7 or the 7th level. I also tried taking this line from my calculations... Sn = 1((2^n)-1) and graphed it as y = 1((2^x)-1). I get the correct values for 'y' (# of students) but the x values are one less then they should be in each case. For example, at x=6, y=63 (one off from the value stated in the question). The answer is so close to being correct but the level is off by one as well as the 'y' value. Is there an explanation to this or something I missed about gemoetric sequences. Any help would be awesome! Thanks a million.

    Also, part c) of the same question asks - 'By the nth level how many students have been contacted? What exactly is the nth level?

    - JDK
     
  2. jcsd
  3. Jan 14, 2004 #2
    Look in Level1 : There is one Element (teacher)
    Level2 : There are 2 Elements = 22-1 (student)
    Level3 : There are 4 Elements = 23-1 (student)
    .....
    .....
    Leveln : There are 2n-1 elements

    So total no of elements upto nth level will be
    1+2+4+......2n-1= 2n-1

    for part c)
    now students contacted by nth level
    would be 2n-2 subtracting one teacher

    For part a)

    2n-2=64 when
    n=6.044
    Now n is an integer so n=6 OR 7

    But for n=6 26-2=62 students would have been contacted so
    n=7 is the level when 64th student have been contacted
     
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