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Homework Help: Formula for the nth term

  1. Jul 28, 2011 #1
    1. The problem statement, all variables and given/known data

    Find a formula for the nth term that determines each sequence.

    2. Relevant equations

    3. The attempt at a solution

    c) 6, 5, 4, 3
    I said this one was n-1 , but apparently its 7-n . can't it be both? How the heck do you tell which one it is?

    f) -3, -6, -12, -24

    I said this one was -2n

    But somehow its -3(2)^n , i dont get how its not -2n, its just being multiplied by 2 .. Anyone?
  2. jcsd
  3. Jul 28, 2011 #2
    I am going to go off of my memory here. Remember the formula for finding a arithmetic sequence is:


    where a1 is the first term of the sequence, an is the nth term of the sequence, and d is the difference between a1 and a2. Whatever the above formula gives you, it should be the answer. It does look like 7-n is the right answer.

    So for the ratio one, the formula is:


    Same definitions as above except r is the ratio between a1 and a2
    Last edited: Jul 28, 2011
  4. Jul 28, 2011 #3
    So, everything esentially has to be in terms of (n-1) ? Genreally speaking, in terms of the subt/addition
  5. Jul 28, 2011 #4
    [STRIKE]Pretty much[/STRIKE]Not always. We can see that a1 = 6, d=-1, so the formula is an=6+(n-1)*-1
    you can do some algebraic manipulations to get 7-n.

    Edit: I take that back. Not everything will be in n+# terms.
    Last edited: Jul 28, 2011
  6. Jul 28, 2011 #5
    but, if you have something like...

    5 , 10 , 15 , 20 ..

    you can't say that this is (n+5)

    but you can say that its 5n ?

    Because 5(n) ?
    Where we are now ignoring the 1 completely?

  7. Jul 28, 2011 #6
    You are right, we can't say it is n+5. Using your first example, we can see that -1 is the difference between all the terms. Let's put your 5,10,15,20 example into my arithmetic sequence formula. We can see that the difference between all terms is 5 and the first term is 5. So using that information we come up with


    an=5+5n-5 (distribution)

    an=5n (cancellation of 5-5)

    I see what you mean by having it in n+5 terms. You are right it won't always be like n+5, but in your original example, it would be in that form.
  8. Jul 28, 2011 #7
    When determining a formula for a sequence, keep in mind that n=1 is the first term unless otherwise specified.

    Now, for {6, 5, 4, 3...} you did observe that each term is smaller than the next by 1. That's a good observation. You suggest "n-1". The best thing to do here is see what n-1 actually gives.

    Beginning with n=1:
    1-1 = 0
    2-1 = 1

    We may as well end that experiment now, seeing that the first two terms of our sequence don't match the given one. However, the correct answer, "7-n", yields:

    7-1 = 6
    7-2 = 5
    7-3 = 4
    7-4 = 3

    Which is exactly what we wanted.

    Your suggestion n-1 may be intuitively not that far off.
    If you consider a recursive sequence (a sequence in which you take the previous term and modify it by something each time to get the next term), you could say:
    [itex] a_n=6 [/itex]

    [itex] a_{n+1}=a_n-1 [/itex]

    Can you say now (forget all the recursive talk) what sequence you get if indeed it is defined as "-2n"?

    Remember, start with n=1...
  9. Jul 28, 2011 #8
    Ok, I understand it now.

    here is a word problem.

    13) A single bacterium divides into two bacteria every 10 minutes. If the same rate of division continue for 2 hours how many bacteria will there be?

    I dont get how to set it up.

    2 hours is 120 minutes, the first sequence kinda goes like 1, 2, 4, 8 , 16

    Going by 2n.

    Its a geometric sequence because its a ratio (division/multi) , but I really dont understand what i am supposed to do. Im supposed to find how many after 120 minutes, but does that mean my sequence goes 10, 20, 30, 40 ?

    And my n is 120?

    ie) a = 1
    r= 2
    n= 120
    tn cant be 120 cause that would be # of bacteria.. anyone?
  10. Jul 28, 2011 #9
  11. Jul 28, 2011 #10
    Your sequence will be an = a1r where a1 is the first term and r is the common ratio. Do you know how to get the common ratio?
    Last edited: Jul 28, 2011
  12. Jul 28, 2011 #11
    ratio is 2
  13. Jul 28, 2011 #12
    If your bacteria divide every ten minutes, how many times will they divide in two hours? That will be the n.

    You have a and r correct. Just remember geometric sequences and series take the form
    [itex] ar^{n-1} [/itex]

    Check it, starting with n=1 and so on, and if it's good, cut right to the desired n.
  14. Jul 28, 2011 #13
    You are not going by 2n. That would look like {2(1), 2(2), 2(3),...} which is {2, 4, 6, ...}

    You are going by [itex] 2a_n [/itex] (each new term multiplied by 2), which is a recursive sequence.
  15. Jul 28, 2011 #14
    Solve the following

    Motion ofa pendulum on the first swing is 50 metres. The length of the arc is 0.97 the previous length on each successive swing. What is the length of the ac after 10 swings?

    what about that one? .. makes 0 sense..

    assuming the ratio is 0.97, initial sequence is 50, next sequence is 50-0.97, etc

    n is 10..

    plugged it in as 50(0.97)^10

    didnt get the correct answer..
  16. Jul 28, 2011 #15
    The second swing is not 50-(.97). it is 50(.97), but you seem to see that by your attempt at the form of the sequence.

    And I'm serious about the form of a geometric sequence. You must put n-1 as the exponent. Try that, check it. Always check the first term, and one or two subsequent terms, and the last term.

    Your sequence doesn't even return 50 as the first term.
  17. Jul 28, 2011 #16
    that gives me 36.8, Which si stilll wrong.

    should be 38.01 .
  18. Jul 28, 2011 #17
    I see
    Last edited: Jul 28, 2011
  19. Jul 28, 2011 #18
    I got the right answer.

    Try again. Make sure you use n-1. So, what should your exponent be when n=10?
  20. Jul 28, 2011 #19
    The new one is not geometric, you even said it, it's arithmetic. That means add or subtract, not multiply.
    Last edited: Jul 28, 2011
  21. Jul 28, 2011 #20
    Okay so, the rows. Something plus one more each time. N gives the one more, because it goes 1 2 3 etc. So, how do u start with 30? 30+n? Hmm... What do you think?
  22. Jul 28, 2011 #21
    Oh i see, okay cool.
  23. Jul 28, 2011 #22
    Is the first term supposed to be negative six? Is that supposed to be -14/3?
  24. Jul 28, 2011 #23
    Find the indicated sum for each geometric series.

    S12 for 1 + 2 + 4 ....

    How do i solve this? I know r= 2 , n = 12-1 .. but nothing else...
  25. Jul 28, 2011 #24
    Okay, I mentioned before this geometric form applies to sequences and series. A series is just a sum of the terms in a sequence. So, again the sequence is written as:
    [itex] ar^{n-1} [/itex]
    a=1 again. Just like bacteria problem.
    [tex] \sum_{n=1}^{12} ar^{n-1}[/tex]
    anyway n=1 goes on the bottom and 12 goes on top to show you add the terms from n=1 to n=12.

    Are you supposed to use the formula for the sum of a geometric series or are you supposed to just add it all up?
    Last edited: Jul 28, 2011
  26. Jul 28, 2011 #25
    You're welcome...
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