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: PS: first 50 positive even integers [#permalink]
06 Mar 2009, 13:23

3

This post received KUDOS

The first 50 even integers will be 2,4,6,8....100. Sum of the first 50 even integers = n(+1) = 50 * 51 = 2550. (Infact this fix is not needed to be explicitly specified. Already implied in the statement sum of the first 50 even integers)

Sum of even integers between 102 to 200 inclusive will be

102 + 104 + 108 + ....200 (there will be 50 terms)

Re: PS: first 50 positive even integers [#permalink]
27 Feb 2011, 21:02

Although it defeats the purpose of the information presented in question, it is much simpler way to find out the sum.

Sum = (Ave) (number of terms)

Ave of an evenly spaced set = (First Term + Last Term) / 2 So, Ave = (102 + 200)/2 = 151

number of terms (inclusive) between an evenly spaced set = ((Last Term - First Term)/Difference between in each term) + 1 So, Number of terms N = ((200 - 98)/2)+1 = 50 (The diff between each of the two even integers is 2. In this particular case, you didn't even need to go thru this calculation since it is obvious that there are 50 even and 50 odd numbers in each 100. But the formula helps for other trickier questions)

Therefore, Sum = 151 * 50 = 7550

hope that helps! _________________

-DK --------------------------------------------------------- If you like what you read then give a Kudos! Diagnostic Test: 620 The past is a guidepost, not a hitching post. ---------------------------------------------------------

Re: PS: first 50 positive even integers [#permalink]
27 Feb 2011, 21:26

1

This post received KUDOS

We are already given the sum of first fifty even numbers and now need to find the sum of next 50. If you notice, each number in this case (sum of next 50) is 100 more than the old series (sum of first 50) and there are 50 such numbers, so the sum will be just the sum of first 50 numbers + 50*100 = 2550+5000 = 7550

Re: PS: first 50 positive even integers [#permalink]
19 Jul 2011, 09:37

The first piece of information is redundant here. We need the sum of even integers from 102 to 200(inclusive). 1. To find the sum in such questions, use the formula:

AVG or AM = Sum/no. of terms

2. The set is an evenly spaced set. i.e., each consecutive number has a constant difference with the prev. (const. diff here is 2, since even integers)

3. For an evenly spaced set, Properties: 1. AM or AVG = (first number + last number)/2 i.e., AM = Mid-term 2. AM = Median Now, we can find Avg. avg. = (102+200)/2 = 302/2 = 151 To find number of terms, Use this formula: No. of terms in a evenly spaced set

= [(Last term - First term)/constant difference] + 1

Re: PS: first 50 positive even integers [#permalink]
20 Jul 2011, 03:11

nikhilsamuel89 wrote:

The first piece of information is redundant here. We need the sum of even integers from 102 to 200(inclusive). 1. To find the sum in such questions, use the formula:

AVG or AM = Sum/no. of terms

2. The set is an evenly spaced set. i.e., each consecutive number has a constant difference with the prev. (const. diff here is 2, since even integers)

3. For an evenly spaced set, Properties: 1. AM or AVG = (first number + last number)/2 i.e., AM = Mid-term 2. AM = Median Now, we can find Avg. avg. = (102+200)/2 = 302/2 = 151 To find number of terms, Use this formula: No. of terms in a evenly spaced set

= [(Last term - First term)/constant difference] + 1

no. of terms = (200-102)/2 +1 = 50

Therefore,

Sum = Avg * no. of terms = 151*50 = 7550

Should this even integer start with 0 ? is 0 an even integer ???? 0 + 2 + 4 .... 98? = 2550?

Re: PS: first 50 positive even integers [#permalink]
20 Jul 2011, 03:15

siddhans wrote:

nikhilsamuel89 wrote:

The first piece of information is redundant here. We need the sum of even integers from 102 to 200(inclusive). 1. To find the sum in such questions, use the formula:

AVG or AM = Sum/no. of terms

2. The set is an evenly spaced set. i.e., each consecutive number has a constant difference with the prev. (const. diff here is 2, since even integers)

3. For an evenly spaced set, Properties: 1. AM or AVG = (first number + last number)/2 i.e., AM = Mid-term 2. AM = Median Now, we can find Avg. avg. = (102+200)/2 = 302/2 = 151 To find number of terms, Use this formula: No. of terms in a evenly spaced set

= [(Last term - First term)/constant difference] + 1

no. of terms = (200-102)/2 +1 = 50

Therefore,

Sum = Avg * no. of terms = 151*50 = 7550

Should this even integer start with 0 ? is 0 an even integer ???? 0 + 2 + 4 .... 98? = 2550?

Yes Zero is even integer

gmatclubot

Re: PS: first 50 positive even integers
[#permalink]
20 Jul 2011, 03:15

This week went in reviewing all the topics that I have covered in my previous study session. I reviewed all the notes that I have made and started reviewing the Quant...

I was checking my phone all day. I wasn’t sure when I would receive the admission decision from Tepper. I received an acceptance from Goizueta in the early morning...

I started running as a cross country team member since highshcool and what’s really awesome about running is that...you never get bored of it! I participated in...