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Integers - Sum

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singhmaharaj Rising GMAT Star
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Integers - Sum Post Tue May 06, 2014 2:43 am
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  • Lap #[LAPCOUNT] ([LAPTIME])
    The sum of the first 50 positive even integers is 2,550. What is the sum of the even integers from 102 to 200, inclusive?

    A) 5,100

    B) 7,550

    C) 10,100

    D) 15,500

    E) 20,100

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    Post Tue May 06, 2014 2:47 am
    singhmaharaj wrote:
    The sum of the first 50 positive even integers is 2,550. What is the sum of the even integers from 102 to 200, inclusive?

    A) 5,100

    B) 7,550

    C) 10,100

    D) 15,500

    E) 20,100
    The sum of the first 50 positive even integers is 2550.
    In other words, 2 + 4 + 6 + 8 + ...+ 98 + 100 = 2550

    We want the sum: 102 + 104 + 106 + . . . 198 + 200
    IMPORTANT: Notice that each term in this sum is 100 GREATER than each term in the first sum.
    In other words, 102 + 104 + 106 + . . . 198 + 200 is the SAME AS...
    (100 + 2) + (100 + 4) + (100 + 6) + ... + (100 + 98) + (100 + 100)
    We can rearrange these terms to get: (100 + 100 + ... + 100 + 100) + (2 + 4 + 6 + 8 + ...+ 98 + 100)

    IMPORTANT: There are 50 100's in the red sum, and we're told that the blue sum = 2550

    So, our sum = 50(100) + 2550
    = 5000 + 2550
    = 7550
    = B

    Cheers,
    Brent

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    Post Tue May 06, 2014 2:48 am
    singhmaharaj wrote:
    The sum of the first 50 positive even integers is 2,550. What is the sum of the even integers from 102 to 200, inclusive?

    A) 5,100

    B) 7,550

    C) 10,100

    D) 15,500

    E) 20,100
    Another approach:

    We want to evaluate: 102 + 104 + 106 + . . . 198 + 200
    Since the terms in this series are EQUALLY SPACED, the sum = (average of first and last term)(# of terms)
    So, sum = [(102 + 200)/2][50]
    = [(302)/2][50]
    = [151][50]
    = 7550
    = B

    Cheers,
    Brent

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    Last edited by Brent@GMATPrepNow on Thu Mar 30, 2017 12:03 pm; edited 1 time in total

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    j_shreyans GMAT Destroyer! Default Avatar
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    Post Tue Feb 17, 2015 9:44 am
    Hi Brent ,

    In this question how did you find the no. of term by which formula?


    Please advise.

    Thanks,

    Shreyans

    Post Tue Feb 17, 2015 10:08 am
    j_shreyans wrote:
    Hi Brent ,

    In this question how did you find the no. of term by which formula?


    Please advise.

    Thanks,

    Shreyans
    Hi Shreyans,

    The most popular rule says: the number of integers from x to y inclusive equals y - x + 1 BUT this applies only to consecutive integers.

    There are other formulas that people like to remember when we're dealing with multiples, but it only takes a couple of seconds to apply the green rule above.

    We have the sequence: 102 + 104 + 106 + . . . 198 + 200
    If we factor out a 2, we get: 2(51 + 52 + 53 + . . . 99 + 100)
    Notice that the number of terms in 102 + 104 + 106 + . . . 198 + 200 is THE SAME AS the number of terms in 51 + 52 + 53 + . . . 99 + 100
    Since 51, 52, 53, ...99, 100 are consecutive integers, we can apply the green rule above.
    The number of terms = 100 - 51 + 1 = 50

    Cheers,
    Brent

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    Post Tue Feb 17, 2015 10:10 am
    Quote:
    The sum of the first 50 positive even integers is 2,550. What is the sum of the even integers from 102 to 200 inclusive?

    a. 5,100
    b. 7,550
    c. 10,100
    d. 15,500
    e. 20,100
    Sum = (number of integers) * (average of biggest and smallest)

    To count the number of evenly spaced integers in a set:

    Number of integers = (Biggest - Smallest)/(distance between each successive pair) + 1

    The distance between each successive pair of even integers is 2.
    Thus, the number of even integers from 102 to 200 = (200-102)/2 + 1 = 50.
    Average of biggest and smallest = (200+102)/2 = 151.
    Sum = (number of integers) * (average of biggest and smallest) = 50*151 = 7550.

    The correct answer is B.

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    Post Tue Feb 17, 2015 6:04 pm
    Hi j_shreyans,

    The required level of 'detail' in these types of questions varies depending on the description of the terms, what you're asked to solve for and how "spread out" the answer choices are.

    In this prompt, the answer choices are so "spaced out" that even if you mis-calculated the number of terms in the sequence, then assuming that the rest of your math was done correctly, you'd end up with an answer that was 'close enough' to the correct one for you to pick it without much concern.

    Also, the first part of the prompt clues you in to how many terms there are:

    The first 50 positive even integers are 2 through 100, inclusive (2x1, 2x2, 2x3 .....2x50). The NEXT 50 terms would have to be from 102 to 200, inclusive, because they're in the same 'format' as the first set of 50 terms (2x51, 2x52.....2x100).

    You could even do a limited set of terms to define the pattern:

    eg.
    The first 5 terms: 2, 4, 6, 8, 10
    The next 5 terms: 12, 14, 16, 18, 20
    Etc.
    Each set of 5 terms takes you "up" another 'multiple of 10'

    So 50 terms = 10 sets of 5 = would take us up to 100
    and 100 terms = 20 sets of 5 = would take us up to 200

    Remember that almost everything that you'll face on Test Day is based on a pattern of some kind. If the 'big pattern' seems difficult, then try dealing with a smaller version of the same pattern and you'll likely learn what you need to deduce the big pattern.

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    Post Thu Mar 30, 2017 11:29 am
    Hey Brent,

    ""So, sum = [(102 + 300)/2][50]""

    You have a typo here. The last term is 200, not 300.

    Thanked by: Brent@GMATPrepNow
    Post Thu Mar 30, 2017 12:04 pm
    deojason wrote:
    Hey Brent,

    ""So, sum = [(102 + 300)/2][50]""

    You have a typo here. The last term is 200, not 300.
    Thanks for catching that.
    I've edited my post.

    Cheers,
    Brent

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