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Math Help - Arithmetic Sequence-What Am I Doing Wrong?

  1. #1
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    Angry Arithmetic Sequence-What Am I Doing Wrong?



    Even if i try simplifying the fraction it doesn't come out to 65/2, which is the correct answer in the back of the book.

    Am i missing something?

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  2. #2
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    \sum_{k=0}^{19} {k-3\over4} = \frac14(\sum_{k=0}^{19} k - \sum_{k=0}^{19} 3) = \frac14({19(19+1)\over2} - 20\cdot3) = \frac{65}2.

    Are you using that \sum_{k=0}^N k = {N(N+1)\over2}?
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  3. #3
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    a(19) = a(1) + (19-1)d
    Now find a(19)
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  4. #4
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    Quote Originally Posted by sa-ri-ga-ma View Post
    a(19) = a(1) + (19-1)d
    Now find a(19)
    Already tried that dude. I still got the wrong answer.

    Quote Originally Posted by maddas View Post
    \sum_{k=0}^{19} {k-3\over4} = \frac14(\sum_{k=0}^{19} k - \sum_{k=0}^{19} 3) = \frac14({19(19+1)\over2} - 20\cdot3) = \frac{65}2.

    Are you using that \sum_{k=0}^N k = {N(N+1)\over2}?
    Man, that totally confused me.

    I know I've seen that formula before, but i can't seem to remember it.
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  5. #5
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    a(19) = -3/4 + 18*1/4 = 15/4.
    So sum is given by
    19/2*(-3/4 + 15/4)
    Simplify and check the answer.
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  6. #6
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    It's still wrong. I get 57/2.
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  7. #7
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    Since your sum starts with a_0, not a_1, the number of terms is 20, not 19.
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  8. #8
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    arithmetric sequence

    Quote Originally Posted by Tazo View Post


    Even if i try simplifying the fraction it doesn't come out to 65/2, which is the correct answer in the back of the book.

    Am i missing something?


    If the first term is -3/4,common diff 1/4 the sum of 20 terms is 65/2. You should write your questions exactly as in your book. Otherwise everyone is confused.


    bjh
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