Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: The sum of the first 50 positive even integers is 2550. What [#permalink]
26 Jan 2013, 10:29

Sachin9 wrote:

use formula

sum = n/2[2a + (n-1)d ]

a first term, n no of terms. d difference..

works like charm :D

To add:

AP Sum = n/2[1st term + last term] which is equivalent to = no. of terms x average _________________

"Appreciation is a wonderful thing. It makes what is excellent in others belong to us as well." ― Voltaire Press Kudos, if I have helped. Thanks! shit-happens-my-journey-to-172475.html#p1372807

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

Re: The sum of the first 50 positive even integers is 2550. What [#permalink]
28 Jan 2013, 10:19

2

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.).

Re: The sum of the first 50 positive even integers is 2550. What [#permalink]
03 Oct 2013, 04:33

Expert's post

Aldossari wrote:

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. _________________

Given the recent news this is way overdue, but see below my first interview report for INSEAD. I met with a senior politician alum who, understandably, had a busy schedule. The...

One thing I did not know when recruiting for the MBA summer internship was the following: just how important prior experience in the function that you're recruiting for...