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Re: The sum of the first 50 positive even integers is 2550. What [#permalink]

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28 Jan 2013, 08:08

From what i see the first part of the question has nothing to do with the problem solving itself..the average between the last term and the first term multiplied by the number of terms will give us the sum of all numbers..number of terms is equal to the average between the last even integer and first even integer plus 1 in this case;(200-102)/2+1=50..therefore the sum of the even integers is equal to 50*(102+200)*1/2=7550 hence B

The sum of the first 50 positive even integers is 2550. What [#permalink]

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28 Jan 2013, 10:19

4

This post received KUDOS

Even easier method:

First 50 even integers are 2, 4, 6, ... , 100. Sum is stated as 2550. Sum of even integers between 102 and 200 are 100 more than each integer in the previous sum (102 is 100 more than 2, etc.).

I am totally confused. in the math book, the formula for the first n even number is n(n+1)

and here there is another formula used, mean*# of terms

Bunuel please explain!!!

The sum of n first positive even numbers is n(n+1) --> the sum of first 50 positive even numbers is 50*51 = 2,550 The sum of evenly spaced set is (mean)*(# of terms) --> the sum of first 50 positive even numbers is (2+100)/2*50 = 51*50 = 2,550.
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Re: The sum of the first 50 positive even integers is 2550. What [#permalink]

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27 Aug 2016, 20:10

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Re: The sum of the first 50 positive even integers is 2550. What [#permalink]

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01 Jul 2017, 14:15

Always ask why the information is provided. This question can be solved without knowing the sum of the first 50 even integers. GMAT tests our ability to catch patterns. Knowing that first 50 even integers add up to 2550 makes life easier here. Rearranging terms we get: 102+104+........+198+200=(100+2)+(100+4)+....+(100+48)+(100+50)=50*100+(2+4+....+48+50) =5000+2550=7550
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Kudosity killed the cat but your kudos can save it.

The sum of the first 50 positive even integers is 2550. 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