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

  • Thread starter YASuperhero
  • Start date
  • Tags
    Sequence
In summary, the equation for finding the nth term in a sequence is n = (3/2)x(x+1), where x is the position of the desired term in the sequence. To find the position of a given term, solve the quadratic equation n^2 + n - 72 = 0 using the given term and 3 as the coefficients.
  • #1
YASuperhero
2
0

Homework Statement


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.


Homework 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!



The Attempt at a Solution


I don't know :(
 
Physics news on Phys.org
  • #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?
 
  • #3
Villyer said:
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?

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?
 
  • #4
YASuperhero said:

Homework Statement


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.

Homework Equations


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

I'll get 108.

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

What's the equation?

Thanks!

The Attempt at a Solution


I don't know :(
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.
 
  • #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!
 
  • #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).
 
  • #7
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:
 

1. How do I find the nth-term of a number in a sequence?

The nth-term of a number in a sequence can be found by determining the pattern or rule of the sequence and using it to find the term at the given position. This can involve using algebraic expressions, arithmetic operations, or geometric formulas.

2. What is the formula for finding the nth-term of a sequence?

The formula for finding the nth-term of a sequence depends on the type of sequence. For arithmetic sequences, the formula is an = a1 + (n-1)d where an is the nth-term, a1 is the first term, and d is the common difference. For geometric sequences, the formula is an = a1rn-1 where an is the nth-term, a1 is the first term, and r is the common ratio.

3. Can I use a calculator to find the nth-term of a sequence?

Yes, you can use a calculator to find the nth-term of a sequence. Most scientific calculators have functions for finding the nth-term of both arithmetic and geometric sequences. However, it is important to understand the formula and concept behind finding the nth-term in order to use the calculator effectively.

4. What if I cannot determine the pattern or rule of a sequence?

If you are unable to determine the pattern or rule of a sequence, you can try looking for differences between terms or ratios between terms. You can also try plotting the terms on a graph to see if there is a visual pattern. If all else fails, you may need to consult a math expert or use a calculator to find the nth-term.

5. How can I use the nth-term to continue a sequence?

The nth-term can be used to continue a sequence by substituting the value of n in the formula to find the corresponding term. For example, if the nth-term of an arithmetic sequence is an = 2n + 3, you can find the 6th term by substituting n = 6 to get a6 = 2(6) + 3 = 15. This process can be repeated for any value of n to continue the sequence.

Similar threads

Replies
3
Views
380
  • Introductory Physics Homework Help
Replies
9
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Precalculus Mathematics Homework Help
Replies
7
Views
1K
Replies
4
Views
811
  • Introductory Physics Homework Help
Replies
20
Views
882
  • Introductory Physics Homework Help
Replies
12
Views
4K
  • Introductory Physics Homework Help
Replies
5
Views
1K
Replies
3
Views
1K
Back
Top