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

  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. 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. 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. HallsofIvy

    HallsofIvy 40,308
    Staff Emeritus
    Science Advisor

    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. 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. 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. AGNuke

    AGNuke 456
    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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