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Math Help - Finding 'a' given two summations

  1. #1
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    Finding 'a' given two summations

    I've been stuck on how to do this question for over an hour. I clearly don't understand how to do it, hopefully somebody can help explain this to me?

    Given:

    \sum\limits_{k = 1}^{100}{(6a_k + 8)} = 2000

    \sum\limits_{j = 0}^{98}{(a_j_+_1)} = 545

    find a100

    Thank you.
    Last edited by Kevmck; November 18th 2012 at 03:14 PM.
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  2. #2
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    Re: Finding 'a' given two summations

    Quote Originally Posted by Kevmck View Post
    I've been stuck on how to do this question for over an hour. I clearly don't understand how to do it, hopefully somebody can help explain this to me?

    Given:

    \sum\limits_{k = 1}^{100}{(6a<sub>k</sub> + 8)} = 2000

    \sum\limits_{j = 0}^{98}{(a<sub>j+1</sub>​)} = 545

    find a100

    Thank you.

    Please edit your post.
    The formatting makes it impossible to know what you mean.
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  3. #3
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    Re: Finding 'a' given two summations

    Given:

    \sum\limits_{k = 1}^{100}{(6a_k + 8)} = 2000

    \sum\limits_{j = 0}^{98}{(a_j_+_1 + 5)} = 545

    find a100​

    Sorry, I wasn't sure how to fix the formatting before, I shouldn't have left it like that until I figured out how. Fixed.
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  4. #4
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    Re: Finding 'a' given two summations

    Quote Originally Posted by Kevmck View Post
    Given:

    \sum\limits_{k = 1}^{100}{(6a_k + 8)} = 2000

    \sum\limits_{j = 0}^{98}{(a_j_+_1 + 5)} = 545

    find a_{100}​

    \sum\limits_{k = 1}^{100} {\left( {6a_k  + 8} \right)}  = 6\sum\limits_{k = 1}^{100} {a_k }  + 8(100)

    \sum\limits_{j = 0}^{98} {\left( {a_{j + 1}  + 5} \right)}  = \sum\limits_{j = 1}^{99} {a_j }  + 5(99)
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  5. #5
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    Re: Finding 'a' given two summations

    I'm not sure I really understand, but thank you for the help.
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  6. #6
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    Re: Finding 'a' given two summations

    Hello, Kevmck!

    \text{Given: }\:(1)\;\sum\limits_{k = 1}^{100}{(6a_k + 8)} \:=\: 2000 \qquad (2)\;\sum\limits_{j = 0}^{98}{(a_j_+_1 + 5)} = 545

    \text{Find }a_{100}.

    (1)\;\sum^{100}_{k=1}(6a_k + 8) \;=\;2000

    . 6\sum^{100}_{k=1} a_k + \sum^{100}_{k=1}8 \;=\;2000

    . . 6\sum^{100}_{k=1}a_k + 800 \;=\;2000

    . . . . . . 6\sum^{100}_{k=1}a_k \;=\;1200

    . . . . . . . \sum^{100}_{k=1}a_k \;=\;200\;\;[3]


    (2) \;\sum^{98}_{j=0}\left(a_{j+1} + 5\right) \;=\;545

    . \sum^{98}_{j=0}a_{j+1} + \sum^{98}_{j=0}5 \;=\;545

    . . \sum^{98}_{j=0}a_{j+1} + 495 \;=\;545

    . . . . . . \sum^{98}_{j=0}a_{j+1} \;=\;50\;\;[4]


    From [3], we have: . a_1 + a_2 + a_3 + \cdots + a_{99} + a_{100} \:=\:200

    From [4], we have: . a_1 + a_2 + a_3 + \cdots + a_{99} \qquad\quad =\;\;50


    Subtract: . a_{100} \:=\:150
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  7. #7
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    Re: Finding 'a' given two summations

    Thank you Soroban. In writing out potential solutions after getting the help from Plato, I eventually came up with 200 and 50 but I didn't know what I was supposed to do with them from there. This example has helped solidify what was happening and what the question was asking for. Thank you again.
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