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sum of all even numbers from 99 to 301

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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by pink_08 » Wed Dec 23, 2009 10:20 am
One small correction above -benjibo
2. Number of even (or odd terms) in a consecutive set = (Last Term - First Term)/2 + 1 [ if the number of elements includes the first term and the last term ] For e.g if the series is 1 2 3 4 5 6 7 8 9

then the number of even numbers is 4 . Using the above formula we would get 5 whereas only 4 even no are present
So we need to add that extra 1 in the formula depending on including the boundary integers



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by valleeny » Wed Dec 23, 2009 10:06 pm
Hi pink_08 for the sequence you gave, to find even numbers in that sequence you have to redefine the set. So for your example it's redefined to 2,3,4,5,6,7,8 essentially. So using the formula you will still get 4 even numbers.
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by benjiboo » Sun Dec 27, 2009 11:12 pm
pink_08 wrote:One small correction above -benjibo
2. Number of even (or odd terms) in a consecutive set = (Last Term - First Term)/2 + 1 [ if the number of elements includes the first term and the last term ] For e.g if the series is 1 2 3 4 5 6 7 8 9

then the number of even numbers is 4 . Using the above formula we would get 5 whereas only 4 even no are present
So we need to add that extra 1 in the formula depending on including the boundary integers



Hi Pink_08. Please read the part that says,

"IMPORTANT: Notice how I used the word "term" and not number. This is important because sometimes you don't always just put the first and last number you are given. For example, If you are asked to find the number of even integers between 1 and 30, you don't use the "first number" in the set. The first number is "1", which is odd, and we are only speaking about even numbers. Therefore, the first term is "2", not "1", even though the set or question might have stated "from 1-30". Same goes with the last term."

This is the same as valleeny was saying. You must redefine the set. "First time", "Last Term".

PM me if you are still confused.
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by mmslf75 » Thu Jan 07, 2010 8:45 am
PERHAPS, another way to solve this is :

Sum of first 150 even integers is N ( N + 1 ) = 150 * 151
Sum of first 48 even interhers is 49 * 50 ... by the same token.

Total = 22650 - 2450 = 20200 ;-)


Sum of first N even is N (N+1)
for odd N^2
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by akuma » Fri Jan 15, 2010 3:16 pm
Sometimes, instead of memorizing rules, go through the problem using smaller numbers a couple times. You'll understand the concept more intuitively.

Let's pick workable numbers 2 to 6, inclusive.
1) sum of ODD numbers => 3+5 = 8
2) sum of EVEN numbers => 2+4+6 = 12

Now let's apply the formula!
AVG of set * (# of terms)

1)
AVG of set = (3+5)/2 = 4
# of terms = (5-3)/2 + 1 = 2
FORMULA = 4*2 = 8

2)
AVG of set = (6+2)/2 = 4
# of terms = (6-2)/2 + 1 = 3
FORMULA = 4*3 = 12
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