sum of n consecutive integers is given, whats n - help pleae

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Source: — Data Sufficiency |

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by GmatKiss » Sat Aug 20, 2011 3:33 am
Please check this post,

https://www.beatthegmat.com/value-of-n-t63911.html

Hope it helps!

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by GmatKiss » Sat Aug 20, 2011 3:38 am
The below compilation may help!!

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

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

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by gmatdriller » Mon Oct 17, 2011 9:19 pm
GmatKiss wrote:The below compilation may help!!

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
First N consecutive Integers
-----------------------------------------------------------------

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 First N Even Consecutive Integers
------------------------------------------------------------------

SUM OF ODD NUMBERS:

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

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
Sum of First N Odd Consecutive Integers

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