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sum of even integers

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Source: — Problem Solving |

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by Anurag@Gurome » Mon Aug 08, 2011 7:02 am
finance wrote:The sum of the first 50 positive even integers is 2550. What is the sum of even integers from 102 to 200 inclusive?
For any set of uniformly spaced integers, the average of the integers = (Largest integer + Smallest integer)/2

In this case, all the even integers between 102 and 200 are uniformly spaced. Hence, average of them = (102 + 200)/2 = 151

Now, sum of all these integers = (Average)*(Number of integers in the range) = 151*[(200 - 102)/2 + 1] = 151*50 = 7550

The correct answer is B.
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by gmatboost » Mon Aug 08, 2011 7:45 am
The "clever approach" which I think is the one used in the OG to explain this is also worth thinking about.

Each number in the bigger list (102, 104, ... 200)
is 100 bigger than the corresponding number in the smaller list (2, 4, ... 100)

So, since there are 50 numbers, the sum of the bigger list is 50*100 = 5000 bigger.
2550 + 5000 = 7550.
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by finance » Mon Aug 08, 2011 7:56 am
gmatboost wrote:The "clever approach" which I think is the one used in the OG to explain this is also worth thinking about.

Each number in the bigger list (102, 104, ... 200)
is 100 bigger than the corresponding number in the smaller list (2, 4, ... 100)

So, since there are 50 numbers, the sum of the bigger list is 50*100 = 5000 bigger.
2550 + 5000 = 7550.
wow I loved this approach...))thank u both guys for ur explanations...I tried to use the first approach but I was not sure whether the rule of" sum of first and last over 2"was for consecutive numbers only or for evenly distributed numbers. Both ways are clear to me now.
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