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Math Help - Sequences - Arithmetic and Recursively Defined

  1. #1
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    Sequences - Arithmetic and Recursively Defined

    There are two problems that are bothering me on my homework. One is this:



    The other is a story problem:

    An architect is asked to design a theater with 20 seats in the first row, 24 in the second row, 28 in the third, and so on. If the theater is to have a capacity of 2480 seats, how many rows must there be?

    Can anyone help with these two?
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  2. #2
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    Quote Originally Posted by blacksuzaku View Post

    An architect is asked to design a theater with 20 seats in the first row, 24 in the second row, 28 in the third, and so on. If the theater is to have a capacity of 2480 seats, how many rows must there be?
    This looks like an arithmetic sequence - remember that in an arithmetic sequence, a common difference is added (or subtracted) each time. Here, we're adding 4 each time. To find the sum of the first n terms of an arithmetic sequence, we use:

    S_{n} = \frac{n(a_{1} + a_{n})}{2}.

    Also, a_{n} = a_{1} + d(n-1) where d is the common difference.

    Combining these, along with the fact that we know a_{1} = 20, the sum, S_{n} = 2480 and d, the common difference, is 4, gives:

    2480 = \frac{n[20 + 20 + 4(n-1)]}{2}.

    Now, solve for n; you'll get a quadratic which factors, although the quadratic formula is probably the easier way to go.
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  3. #3
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    First bit...

    a_2 = \frac{2(2^3) + 2}{3(2^2)} = \frac{3}{2}

    Then reinsert 3/2 to get the next one, etc...
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  4. #4
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    I did that, Sean... but when I hit a6, I ended up with a number that had an insane amount of digits. ._.;;
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  5. #5
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    Keep it in fraction form.
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  6. #6
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    Hello, blacksuzaku!

    Here's the second one . . .


    An architect is asked to design a theater with 20 seats in the first row,
    24 in the second row, 28 in the third, and so on. If the theater is to have
    a capacity of 2480 seats, how many rows must there be?

    The sum of the first n terms of an arithmetic series: . S_n \;=\;\frac{n}{2}\bigg[2a + (n-1)d\bigg]

    . . where: . a = first term, d = common difference, n = number of terms.


    We have an arithmetic series: . S_n \;=\;20 + 24 + 28 + 32 + \hdots\;=\;2480

    It has: . a = 20,\;\;d = 4,\;\;n\text{ terms}

    So we have: . \frac{n}{2}\bigg[2(20) + (n-1)4\bigg] \:=\:2480

    . . which simplifies to: . n^2 + 9n - 1240 \:=\:0

    . . which factors: . (n-31)(n+40) \:=\:0

    . . and has the positive root: . n \:=\:31


    Therefore, the theater will have 31 rows.

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  7. #7
    Behold, the power of SARDINES!
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    Quote Originally Posted by blacksuzaku View Post
    There are two problems that are bothering me on my homework. One is this:



    The other is a story problem:

    An architect is asked to design a theater with 20 seats in the first row, 24 in the second row, 28 in the third, and so on. If the theater is to have a capacity of 2480 seats, how many rows must there be?

    Can anyone help with these two?
    For the first one it is just plug and chug... for n=2 we get

    a_2=\frac{2a_1^3+2}{3a_1^2}=

    Since we a_1=2 we can just plug it in to get

    a_2=\frac{2(2)^3+2}{3(2)^2}=\frac{18}{12}=\frac{3}  {2}

    Now that we know a_2=\frac{3}{2} we can find the next term... just rinse and repeat...

    For the 2nd we are dealing with an arithmetic series and we will need the formula

    S_n=\frac{n(2a_1+(n-1)d)}{2} where a_1 is the first term and d is the difference between sequence members

    So for us we know that S_n=2480 \\\ a_1=20 \\\ d=4

    So plugging all of this in we get

    2480=\frac{n(2(20)+(n-1)4)}{2} \iff 4960=40n+4n^2-4n \iff 4n^2+36n-4960=0

    Now factoring a bit we get

    4[n^2+9n-1240]=0 \iff 4(n+40)(n-31) so we get n=-40 and n=31.

    Since we can't build a negative number of rows we get that we need 31 rows.

    I hope this helps.

    Good luck.

    Edit: Too slow. sigh
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  8. #8
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    Let's just say that the answer I got for a5 was 2935497269576521/2329904227757400

    And I still have to plug that into a6... see my problem?
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  9. #9
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    I suggest you use the Ans button on your calculator to refer to the previous term and give each term to 3 significant figures.
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  10. #10
    Behold, the power of SARDINES!
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    Spred sheet

    I ran it in excel with 12 iterations and this is what I came up with

    Sequences - Arithmetic and Recursively Defined-capture1.jpg
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  11. #11
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    Well that merits the proverbial roflcopter... I feel stupid now :P

    Stupid TI-89 and its "ZOMG lets keep everything in rational form"...
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