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How to Find what nthterm a Number is in a Sequence 
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#1
Jun1912, 07:16 PM

P: 2

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
Jun1912, 07:40 PM

P: 294

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? 


#3
Jun1912, 08:02 PM

P: 2

What am I missing? 


#4
Jun1912, 08:15 PM

Math
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Thanks
PF Gold
P: 39,682

How to Find what nthterm a Number is in a Sequence
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. 


#5
Jun1912, 08:57 PM

P: 546

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! 


#6
Jun1912, 09:10 PM

P: 294

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). 


#7
Jun2012, 02:09 AM

PF Gold
P: 456

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 


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