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Formula for consecutive, even, odd integers

This topic has 3 expert replies and 9 member replies
California4jx Master | Next Rank: 500 Posts Default Avatar
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Formula for consecutive, even, odd integers

Post Wed Aug 27, 2008 9:44 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    Thought to put together all formulas at one place:


    Sum of all consecutive integers:

    N (N+1) / 2 ------- (I)
    where N is the number of terms.

    Example: 1+2+3+........ + 100
    N=100
    (I) => 100 (101) / 2
    => 50 (101) = 5050

    -----------------------------------------------------------------

    SUM OF EVEN NUMBERS:

    Formula: N(N+1)
    How to Find N = (First Even + Last Even)/2 - 1

    Example: 2+4+6+ ....... 100
    N = (2+100)/2 - 1 = 50

    Sum of first 50 positive even integers = (50)(51) = 2550
    ------------------------------------------------------------------

    SUM OF ODD NUMBERS:

    If N = number of odd terms then sum = (N)^2

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    rishi235 Senior | Next Rank: 100 Posts Default Avatar
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    Post Wed Aug 27, 2008 9:53 am
    Nice work...
    Just to add to ur collection:

    Sum of squares of 1st n consecutive natural nos = n(n+1)(2n+1) / 6

    Sum of cubes of the 1st n consecutive natural nos. = [n(n+1)/2]^2

    California4jx Master | Next Rank: 500 Posts Default Avatar
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    Post Wed Aug 27, 2008 1:09 pm
    Great .. !

    rixyroxy Newbie | Next Rank: 10 Posts Default Avatar
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    Post Fri Aug 19, 2011 10:01 am
    Are these formulae tested on GMAT ?
    rishi235 wrote:
    Nice work...
    Just to add to ur collection:

    Sum of squares of 1st n consecutive natural nos = n(n+1)(2n+1) / 6

    Sum of cubes of the 1st n consecutive natural nos. = [n(n+1)/2]^2

    kapilhede17 Junior | Next Rank: 30 Posts Default Avatar
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    Post Fri Jul 13, 2012 9:03 pm
    I just remember the generic formula

    Sn = n/2( 2a + (n-1)d)
    a=1st term
    d= common diff

    It saves me from remembering the formula , and all the ones mentioned are like a substitution away

    Thanked by: aalexjacob
    nthiru100 Newbie | Next Rank: 10 Posts Default Avatar
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    Post Mon Nov 05, 2012 8:07 am
    Thank You Guys

    Mystiqan Newbie | Next Rank: 10 Posts
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    Post Sat Jun 21, 2014 9:40 am
    Or for even:[x(x+2)]/4; and odd:[x(x+2)+1]/4...x is the greatest digit of your set.ex: out of 7,5,3,1 your x would be 7. Hope this helps. I take credit for figuring these out without any help other than referencing the first equation.

    kiranrai.ch Newbie | Next Rank: 10 Posts
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    Post Sun Jan 10, 2016 9:08 pm
    sum of even number: n((first even+last even)/2)

    where, n = total number of even numbers

    But, you can't use the Formula: N(N+1)
    How to Find N = (First Even + Last Even)/2 - 1

    For example, let's find the sum of even numbers between 102 to 200.

    =50((102+200)/2)
    =7550, which is right.

    chetan.sharma Senior | Next Rank: 100 Posts Default Avatar
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    Post Fri Jan 15, 2016 5:59 am
    kiranrai.ch wrote:
    sum of even number: n((first even+last even)/2)

    where, n = total number of even numbers

    But, you can't use the Formula: N(N+1)
    How to Find N = (First Even + Last Even)/2 - 1

    For example, let's find the sum of even numbers between 102 to 200.

    =50((102+200)/2)
    =7550, which is right.
    hi,
    the formula n(n+1) is for first n even positive integer..
    and will not be true for a set of consecutive even integers picked up from between..

    GMAT/MBA Expert

    Post Fri Feb 19, 2016 2:20 pm
    An easy way to remember the sum of consecutive integers:

    Suppose our set is [n - x, n - (x - 1), n - (x - 2), ...., n, .... n + (x - 2), n + (x - 1), n + x].

    Now let's add terms that are equidistant from the mean:

    (n - x) + (n + x) = 2n

    (n - (x - 1)) + (n + (x - 1)) = 2n

    etc.

    Wow! So every pair of terms that is equidistant from the mean has a sum of 2n, and an average of n.

    Since the sum of any set = (# of terms) * average, we know that our sum = (# of terms) * n.

    From there, just count the number of terms in the set, and you're done.

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    Post Sun Oct 22, 2017 5:35 am
    SUM OF EVEN NUMBERS:

    Formula: N(N+1)
    How to Find N = (First Even + Last Even)/2 - 1

    Example: 2+4+6+ ....... 100
    N = (2+100)/2 - 1 = 50

    Sum of first 50 positive even integers = (50)(51) = 2550

    with the above formula....I tried to solve this question but I got it wrong
    For any positive integer n, the sum of the first n positive integers equals n(n+1)/2. What is the sum of all the even integers between 99 and 301?

    What is the issue?

    Post Sun Oct 22, 2017 1:23 pm
    Two more that come to mind:

    1. The number of integers from x to y inclusive = y - x + 1
    example: the number of integers from 13 to 41 inclusive = 41 - 13 + 1 = 29

    2. The sum of the first k ODD integers = k²
    example: the sum of the first 10 ODD integers = 10² = 100
    That is, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 100

    Cheers,
    Brent

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    GMAT/MBA Expert

    Post Thu Nov 09, 2017 6:12 pm
    I wouldn't do too much with this thread, personally. The GMAT is all about thinking creatively, not formulaically, and the testwriters devise problems to test your ability to handle subtle deviations and curveballs from the normal, boring math drills. On the GMAT, if you don't understand where a formula comes from and how to recreate it (not just recite it!) on your own, you're extremely vulnerable, no matter what topic's best tested. I suppose it's fine to just memorize a$$^2$$ + b$$^2$$ = c$$^2$$, but these "formulas" are unlikely to be tested crudely enough for rote memorization to help you.

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