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Math Help - Find the sum of the integers...

  1. #1
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    Find the sum of the integers...

    Hi

    My problem is:

    Find the sum of the integers between 200 and 499 that are multiples of three.

    I have the answer:

    34,950

    but I was wondering if there was an easier way to get to that besides adding all of the multiples of three? I kept getting messed up that way. Thank you.
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  2. #2
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    Quote Originally Posted by Averee View Post
    Hi

    My problem is:

    Find the sum of the integers between 200 and 499 that are multiples of three.
    .
    List them,

    201,204,207,...,498

    This is arithmetic sequence we commond difference 3.
    Thus,
    a_n=3n+b
    Where, n is the "n-th" term.

    Thus,
    a_1=3(1)+b
    But,
    a_1=201
    Thus,
    201=3+b
    And hence,
    b=198

    In general this sequence can be modelled as,
    a_n=3n+198

    Let us find what "n" is in the last term,
    498=3n+198
    300=3n
    n=100

    Thus, there are 100 terms.
    Hence,
    a_n=3n+198 where "n" ranges from 1 to 100.

    We need to find the sum,
    a_1+a_2+...+a_100
    That is,
    3(1)+198+3(2)+198+3(3)+198+...+3(100)+198
    Rewrite this as,
    3(1+2+...+100)+(198+198+...+198)
    Where 198 appears 100 terms.
    Thus,
    3(1+2+...+100)+100*198

    Now use the formula,
    1+2+...+n=n(n+1)/2

    Hence,

    3(5050)+100*198
    Is the answer.
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  3. #3
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    Thank you. I think I understand most of what you did, except your answer :
    3(5050) + 100 * 198 equals 3,019,500, and my answer is supposed to be 34,950?
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  4. #4
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    Quote Originally Posted by Averee View Post
    Thank you. I think I understand most of what you did, except your answer :
    3(5050) + 100 * 198 equals 3,019,500, and my answer is supposed to be 34,950?
    You made an arithmetical error.
    My answer is correct and it matches with thine.
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  5. #5
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    Okay, what did I do wrong?
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  6. #6
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    Quote Originally Posted by Averee View Post
    Okay, what did I do wrong?
    I think you though that * was a comma. It means times.

    100*198=19800

    3(5050)=15150

    15150+19800=34950
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  7. #7
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    I see, I was doing the order of operations wrong. Thanks again. One more question though (promise), how did you get to that very last part? The one that gave the answer?
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  8. #8
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    Quote Originally Posted by Averee View Post
    I see, I was doing the order of operations wrong. Thanks again. One more question though (promise), how did you get to that very last part? The one that gave the answer?
    What part.
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  9. #9
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    I think I got most of it, but when you put 3(5050) + 100 * 198, how did you know to do that? Did you use one of the formulas you gave me?
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  10. #10
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    Quote Originally Posted by Averee View Post
    I think I got most of it, but when you put 3(5050) + 100 * 198, how did you know to do that? Did you use one of the formulas you gave me?
    Yes.

    1+2+3+...+100 = 101(100)/2=5050

    By using the formula,

    1+2+...+n=(n+1)n/2
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  11. #11
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    Thanks again.
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