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rachelk09

3, 12, 21, 30, 39.... What is the arithmetic function of this sequence? I know we add 9 to it each time but I need to figure out the function. Kind of stuck...

  • one year ago
  • one year ago

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  1. AccessDenied
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    What are your input values of your function, and what are your output values?

    • one year ago
  2. rachelk09
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    I don't have any input values. The output values are as I indicated...3, 12, 21, 30, 39...etc.

    • one year ago
  3. phi
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    add 9 to it each time what do you do to 3 to get each number? add 9, then 2*9 , then 3*9...

    • one year ago
  4. rachelk09
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    If I do f(n) = 3+9n this would work but only if things start at n=0

    • one year ago
  5. AccessDenied
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    For the function, if we want it to represent this sequence, it probably should return the first value as 3 ( f(1) = 3 ), the next value as 12 ( f(2) = 12 ), then 21 ( f(3) = 21 ), and so on. So, these are our 'points.' We just have to figure out the general formula. Use the fact that you add 9 a different number of times to 3 to get to all of your function's output values...

    • one year ago
  6. rachelk09
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    Doesn't the sequence start at n=1

    • one year ago
  7. AccessDenied
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    Hm.. I think you could technically start it at either n=0 or n=1. It shouldn't matter too much, but I'm not sure if your teacher may have a preference that he/she used?

    • one year ago
  8. rachelk09
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    She really didn't say...hmm..I guess I'll just try it and indicate that n starts at 0

    • one year ago
  9. AccessDenied
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    K. If you wanted to move it to n=1 => f(1) = 3, you could just use a function shift: f(n) = 3 + 9n f(n-1) = 3 + 9(n-1) = 3 + 9n - 9 = -6 + 9n <-- call this the new f(n)

    • one year ago
  10. AccessDenied
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    I think I would prefer the start at n=0, just because it is easier to see your first term. :)

    • one year ago
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