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Math Help - Arithmetic series - stuck!

  1. #1
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    Arithmetic series - stuck!

    Hi all, just joined and introduced myself in the lobby.

    I have two questions which I have not been able to do the workings out for. I understand the formula to calculate arithmetic series, but I am stumped on these - hope you can help.

    1. The sum of an arithmetic series is 270. The common difference is 1 and the first term is 4. Calculate the number of terms in the series.

    2. The first term of an arithmetic series is 16. The 30th term is 100. Calculate S30.

    Can anybody show me the process they would go through?

    Thanks

    Parad80.
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  2. #2
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    Re: Arithmetic series - stuck!

    sum of an arithmetic series ...

    S_n = a_1 + (n-1)d

    you know S_n , d, and a_1 ... so, what's the problem with finding n ?
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  3. #3
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    Re: Arithmetic series - stuck!

    Hello, parad80!

    You say you understand the formulas.
    I don't understand your difficulty.


    1. The sum of an arithmetic series is 270.
    The common difference is 1 and the first term is 4.
    Calculate the number of terms in the series.

    The sum is the first n terms of an arithmetic series: . S_n \:=\:\frac{n}{2}\big[2a_1 + (n-1)d\big]

    We are given: . S_n = 270,\;a_1=4,\;d = 1

    We have: . \frac{n}{2}\big[2(4) + (n-1)1\big] \:=\:270

    \text{Solve for (positive) }n.




    2. The first term of an arithmetic series is 16.
    The 30th term is 100.
    Calculate S
    30

    \text{The }n{th}\text{ term is: }\:a_n \:=\:a_1 + (n-1)d

    We are given: . a_1 = 16,\;n=30,\;a_{30} = 100

    We have: . 16 + 29d \:=\:100 \;\hdots\;\text{Solve for }d.

    \text{Then determine }S_{30}
    Thanks from skeeter
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  4. #4
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    Re: Arithmetic series - stuck!

    Quote Originally Posted by skeeter View Post
    sum of an arithmetic series ...

    S_n = a_1 + (n-1)d ... nope, thinking sum, putting down nth term. no more drinking and deriving.
    mea culpa.
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  5. #5
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    Re: Arithmetic series - stuck!

    Thanks for your responses:

    Don't know how you write your formula's in that font, but I understand it up to this point:

    1.
    n/2(8+(n-1)1),

    n/2(8+(n-1)) = 270

    Then what is the next step?

    Thanks for your patience!

    PD
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  6. #6
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    Re: Arithmetic series - stuck!

    expand out the brackets and use the quadratic formula.
     \frac{n}{2}(8+(n-1)) = 270
    n(8+(n-1)) = 540
    8n+ n(n-1) = 540
    n^2 + 7n - 540 = 0
    now use the quadratic formula

    remember the quadratic formula will give you two solutions, only the positive one is a feasible answer.
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  7. #7
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    Re: Arithmetic series - stuck!

    thanks Iinalg123

    understand up to 7n+n2=540

    couldn't you then divide by 7 on both sides?

    With n2 + 7n - 540 is there an easy way to work the quadratic formula out?

    Can someone please tell me how to write the formula's in the fonts above - thanks.

    PD
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  8. #8
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    Re: Arithmetic series - stuck!

    Lets say you have a quadratic ax^2 + bx + c = 0

    the quadratic formula says the solutions are x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

    If you use this formula for your quadratic you will get your solution for x.
    Thanks from parad80
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  9. #9
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    Re: Arithmetic series - stuck!

    ok, so I have the following:

    -7 + SQRT 7 squared - 4(1)(540)
    divided by
    2(1)

    Then:
    -7 + SQRT 49 -2160
    divided by
    2

    Then:
    -7 + SQRT - 2111
    divided by
    2

    Then:
    -7 + 45.95
    divided by
    2

    Then:
    38.95
    divided by
    2

    Then:
    19.475, which rounded up is 20, which is the text book answer - thank you.
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  10. #10
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    Re: Arithmetic series - stuck!

    Close, however you didn't convert it to the form ax^2+bx+c=0, you used ax^2 + bx = c
    So, if we convert it first we have n^2 + 7n -540 = 0 (note the negative sign for c)

    So,  n =  \frac{-7 \pm \sqrt{7^2 - 4(1)(-540)}}{2(1)}

    i.e  n =  \frac{-7 \pm \sqrt{49 +2160)}}{2} (note the sign chance as negative * negative is positive.)

     n =  \frac{-7 \pm \sqrt{2209}}{2}

     n =  \frac{-7 \pm 47}{2}

    now as we only considering the positive answer  n =  \frac{40}{2} = 20
    Thanks from parad80
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  11. #11
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    Re: Arithmetic series - stuck!

    With regard to the second question:

    The first term of an arithmetic series is 16. The 30th term is 100. Calculate S30
    I have the following formula:

    S_n \:=\:\frac{n}{2}\big[2a + (n-1)d\big]

    so

    S_1 \:=\:\frac{1}{2}\big[2(16) + (1-1)d\big] \:=\:\frac{1}{2}\big[32 + d\big] \:=\:\big[16 + \frac{1}{2}d\big]

    and

    S_3_0 \:=\:\frac{30}{2}\big[2(100) + (30-1)d\big] \:=\:\frac{30}{2}\big[200 + 29d\big] \:=\:\15\big[200 + 29d\big] \:=\:\big[3000 + 435d\big]

    Where from here? Can't help feeling I am going about this in the wrong way.

    PD
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  12. #12
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    Re: Arithmetic series - stuck!

    Close, however you didn't convert it to the form. If we convert it first we have (note the negative sign for c)

    i.e (note the sign chance as negative * negative is positive.)
    Got it - thanks.
    Last edited by parad80; May 29th 2012 at 11:04 PM.
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