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 500,000 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: AP series , need quick way to solve this [#permalink]

Show Tags

27 Apr 2008, 20:19

Even numbers from 102 to 200 = 2* ( 51 to 100 ). No of elements in this series = (100-51+1) = 50. The middle value of this series = (51+100) / 2 = 75.5 The sum = 2 * 75.5 * 50 = 7550.

Re: AP series , need quick way to solve this [#permalink]

Show Tags

27 Apr 2008, 22:11

bsd_lover wrote:

Even numbers from 102 to 200 = 2* ( 51 to 100 ). No of elements in this series = (100-51+1) = 50. The middle value of this series = (51+100) / 2 = 75.5 The sum = 2 * 75.5 * 50 = 7550.

Re: AP series , need quick way to solve this [#permalink]

Show Tags

27 Apr 2008, 22:18

Hi, this is not really "my" method. Its explained really well in the Manhattan GMAT series.

Say if are finding out the sum of a series lets use 1 - 99 for example.

First thing to find out is how many items in the series. In this case - there are 99 items.

Next thing to find out is what is the average of all the items - this is the absolute middle value. This can be found simply by taking the average of first and last elements : 99+1 / 2 = 50 <-- obviously the middle value.

So now we have an average (i.e. a mean) and a number of elements. To find the sum we simply multiply the mean with the number of elements : 50 * 99 = 4950

Re: AP series , need quick way to solve this [#permalink]

Show Tags

27 Apr 2008, 22:29

bsd_lover wrote:

Hi, this is not really "my" method. Its explained really well in the Manhattan GMAT series.

Say if are finding out the sum of a series lets use 1 - 99 for example.

First thing to find out is how many items in the series. In this case - there are 99 items.

Next thing to find out is what is the average of all the items - this is the absolute middle value. This can be found simply by taking the average of first and last elements : 99+1 / 2 = 50 <-- obviously the middle value.

So now we have an average (i.e. a mean) and a number of elements. To find the sum we simply multiply the mean with the number of elements : 50 * 99 = 4950

cheers.

brilliant! that cleared things up very nicely.

does this have anything to do w/ the question at hand: The sum of the first 50 positive even integers is 2,550.

Re: AP series , need quick way to solve this [#permalink]

Show Tags

29 Apr 2008, 02:20

droopy57 wrote:

brilliant! that cleared things up very nicely. does this have anything to do w/ the question at hand: The sum of the first 50 positive even integers is 2,550.

2,550 can be used for another way of solving: the sum of first 50 positive inegers looks like 2 + 4 + 6 + .. + 100, all in all 50 items the sum of next 50 positive integers looks like 2+100 + 4+100 + 6+100 + ... + 100+100, so you can find the sum of even integers from 102 to 200 by adding: 2550 + 100*50 = 7550

Re: AP series , need quick way to solve this [#permalink]

Show Tags

29 Apr 2008, 10:51

Sunchaser20 wrote:

droopy57 wrote:

brilliant! that cleared things up very nicely. does this have anything to do w/ the question at hand: The sum of the first 50 positive even integers is 2,550.

2,550 can be used for another way of solving: the sum of first 50 positive inegers looks like 2 + 4 + 6 + .. + 100, all in all 50 items the sum of next 50 positive integers looks like 2+100 + 4+100 + 6+100 + ... + 100+100, so you can find the sum of even integers from 102 to 200 by adding: 2550 + 100*50 = 7550

Re: AP series , need quick way to solve this [#permalink]

Show Tags

29 Apr 2008, 20:19

bsd,

Why do you use multiplier of 2 in the problem? Yet in the example you provide there is no multiplier. I understand how you derive the 2 using property associations; however, if there are fifty elements in the problem why is the answer not 50*75.5?

Re: AP series , need quick way to solve this [#permalink]

Show Tags

29 Apr 2008, 20:32

Its because the original question deals with a sequence of EVEN integers - which I convert to a regular sequence of integers by introducing the 2.

My second example is generic and uses a regular sequence.

smginnis wrote:

bsd,

Why do you use multiplier of 2 in the problem? Yet in the example you provide there is no multiplier. I understand how you derive the 2 using property associations; however, if there are fifty elements in the problem why is the answer not 50*75.5?

gmatclubot

Re: AP series , need quick way to solve this
[#permalink]
29 Apr 2008, 20:32