How to Find what nth-term a Number is in a Sequence

  1. Jun 19, 2012 #1
    1. The problem statement, all variables and given/known data
    I'm trying to find out the equation for how to find out where a number falls in a sequence.

    An example sequence would be 3, 9, 18, 30, 45, 63, 84, 108, 135, 165...
    108 is the 8th number in that sequence.


    2. Relevant equations
    If I use the equation (xn+xn^2)/2
    (x = 3 & n = 8)

    I'll get 108.

    What I need is to use the 108 and 3 and get 8.

    What's the equation?

    Thanks!



    3. The attempt at a solution
    I dont know :(
     
  2. jcsd
  3. Jun 19, 2012 #2
    Give the nth term of the sequence a variable (anything works, you just can't reuse x or n).

    Then you get a = [itex] \frac{xn+xn^{2}}{2} [/itex], and you are looking for a function of a and x for n. Does this help at all?
     
  4. Jun 19, 2012 #3
    I don't understand. That is the equation I have and that will give me a. But I have a and x. What I need to find out is n.

    What am I missing?
     
  5. Jun 19, 2012 #4

    HallsofIvy

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    That's NOT an equation. I think you mean [itex]x_{n+1}= (x_n+ x_n^2)/2[/itex].
    I don't know what you mean by "x= 3 & n= 8". There is NO "x" in the formula. If you start counting with 1 (that is, [itex]x_1= 3[/itex] then [itex]x_8= 108[/itex] alright but I don't know where the "x= 3" comes into it.

    In any case, while [itex](x_1+ x_1^2)/2= (3+ 9)/2= 6[/itex] so "[itex]x_{n+1}= (x_n+ x_n^2)/2[/itex] has nothing to do with this sequence.

    Using a "difference" method, I see that the "first differences" ([itex]x_{n+1}- x_n[/itex]) are 9- 3= 6, 18- 9= 9, 30- 18= 12, 45- 30= 15, etc. and the "second differences" are 9- 6= 3, 12- 9= 3, 15- 12= 3, etc. As far as we can see (of course, just seeing some numbers in a sequence doesn't guarentee the sequence will continue in the same way) the "second difference" will always be 3 and by "Newton's difference method" the sequence is produced by the equation [itex]x_n= (3/2)n(n+1)[/itex]. It's easy to check that this is correct:
    (3/2)(1)(1+ 1)= 3, (3/2)(2)(2+ 1) = 9, (3/2)(3)(3+1)= 18, (3/2)(4)(4+1)= 30, and so on.

    You can't just "use the 108 and 3" to get 8 because the sequence depends on more than that. Once you know that [itex]x_n= (3/2)n(n+1)[/itex] you just need to solve (3/2)n(n+1)= 108 for n. From that, n(n+1)= 108(2/3)= 2(36)= 72. You could solve the quadratic equation [itex]n^2+ n- 72= 0[/itex] by factoring or using the quadratice equation. But knowing that n must be an integer, it is easier to note that n and n+2 are not far so seeing that the square root of 72 is between 8 and 9 we would "guess" than n= 8 and n+1= 9.
     
  6. Jun 19, 2012 #5
    Subtract the difference between all of the numbers notice that there is another pattern:
    6,9,12,15,18, etc....

    Taking the difference again you notice that they are all 3 apart.

    I can tell you that based on this information the equation which describes these numbers is quadratic. You can derive the coefficients using this information, good luck!
     
  7. Jun 19, 2012 #6
    Hallsofivy, his equation is based on two variables, x and n. His choosen x is 3.

    That's why [itex]\frac{3}{2}n(n+1)[/itex] is a representation of his sequence.


    The way I see it is that a(x,n) is the nth term of of the sequence [itex]\frac{x}{2}n(n+1)[/itex], and he is looking for a function n=(a,x).
     
  8. Jun 20, 2012 #7

    AGNuke

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    Gold Member

    No doubt that this equation is a quadratic, as explained by Aero51.

    Now we can take a general form of equation as [itex]ax^2+bx+c[/itex]

    Now for x = 1, a(1) + b(1) + c = 3 => a + b + c = 3
    For x = 2, a(4) + b(2) + c = 9 => 4a + 2b + c = 9
    For x = 3, a(9) + b(3) + c = 18 => 9a + 3b + c = 18

    Now solve these 3 equations to get a, b, c and your equation :wink:
     
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