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:

For any positive integer n, the sum of the first n positive integers equals n(n+1)/2. What is the sum of all the even integers between 99 and 301?

Answer is B: 20,200.

Don't understand the explanation - somehow they applied the above formula and customized it to even integers but I don't really follow. Can anyone explain the best way to solve? Thanks!

In any set of consecutive integers the sum = average*number of integers

average = 200 total = 150-50 + 1 = 101

150 because 300 is the 150th even integer 50 because 100 is the 50th even integer

For any positive integer n, the sum of the first n positive integers equals n(n+1)/2. What is the sum of all the even integers between 99 and 301?

Answer is B: 20,200.

Don't understand the explanation - somehow they applied the above formula and customized it to even integers but I don't really follow. Can anyone explain the best way to solve? Thanks!

sum of all even integer 1 to 301 = 2 * (1+2+...+150) = (2*150*151)/2 = 150*151 sum of all even integer 1 to 99 = 2 * (1+2+....+49) = (2*49*50)/2 = 49*50

Schools: Harvard, Stanford, Wharton, MIT & HKS (Government)

WE 1: Economic research

WE 2: Banking

WE 3: Government: Foreign Trade and SMEs

For any positive integer n, the sum of the first n positive integers [#permalink]

Show Tags

22 Oct 2010, 10:22

For any positive integer n, the sum of the first n positive integers equals \((n*(n+1))/2\). What is the sum of all the even integers between 99 and 301?

Re: For any positive integer n, the sum of the first n positive integers [#permalink]

Show Tags

22 Oct 2010, 16:03

metallicafan wrote:

For any positive integer n, the sum of the first n positive integers equals \((n*(n+1))/2\). What is the sum of all the even integers between 99 and 301? (A) 10,100 (B) 20,200 (C) 22,650 (D) 40,200 (E) 45,150

Sum = 100+102+...+300 = 100*101 + (0+2+4+...+200) [Taking 100 from each term in series, there are 101 terms] = 100*101 + 2*(1+2+..+100) [Taking 2 common] = 100*101 + 2*100*101/2 [Using formula given] = 2*100*101 = 20200

Re: For any positive integer n, the sum of the first n positive integers [#permalink]

Show Tags

01 Jan 2011, 20:12

#157: For any positive integer n, the sum of the first n positive integers equals [n(n+1)]/2. What is the sum of all the even integers between 99 and 301? A) 10,100 B) 20,200 C) 22,650 D) 40,200 E) 45,150

Big kudos to someone who can get this right and explain the answer for me!

If there are 100 even integers in the sequence why do you +1 and multiply the average (200) by 101?

Question: even integers between 99 and 301

We know 99 is not even and 301 is not even; so ignore both of those

The actual sequence becomes:

100,102,104,106,108,.......,296,298,300

In the above sequence; the first number 100 is divisible by 2(or is even) and the last number 300 is also divisible by 2(or is even).

The common difference "d" is 2.

For such sequences; there is a formula to find the number of elements:

It is {(Last Number - First Number)/d} + 1

So; the above sequence actually contains {(300-200)/2}+1 = 101 elements

And Average = (first number + last number)/2 = (100+300)/2 = 200

Sum = Average * Number of Elements = 101 * 200 = 20200

Ans: 20200

Your answer is correct, but the red text above contains an error. The last number is 300 and the first number is 100 therefore there are 101 elements. I think your work above should read {(300-100)/2}+1 = 101

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

For any positive integer n, the sum of the first n positive integers equals n(n+1)/2. What is the sum of all the even integers between 99 and 301?

A. 10,100 B. 20,200 C. 22,650 D. 40,200 E. 45,150

Approach #1: Even integer between 99 and 301 represent evenly spaced set (aka arithmetic progression): 100, 102, 104, ..., 300. Now, the sum of the elements in any evenly spaced set is the mean (average) multiplied by the number of terms. (Check Number Theory chapter of Math Book for more: math-number-theory-88376.html)

Average of the set: (largest+smallest)/2=(300+100)/2=200; # of terms: (largest-smallest)/2+1=(300-100)/2+1=101 (check this: totally-basic-94862.html#p730075);

The sum = 200*101= 20,200.

Answer: B.

Approach #2: Using the formula of the sum of the first n positive integers: n(n+1)/2.

100+102+...+300=2(50+51+..+150). Now, the sum of the integers from 50 to 150, inclusive equals to the sum of the integers from 1 to 150, inclusive minus the sum of the integers from 1 to 49, inclusive. 2(50+51+..+150)=2*(150(150+1)/2-49(49+1)/2)=20,200.

Re: For any positive integer n, the sum of the first n positive integers [#permalink]

Show Tags

24 Apr 2016, 13:03

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Version 8.1 of the WordPress for Android app is now available, with some great enhancements to publishing: background media uploading. Adding images to a post or page? Now...

“Keep your head down, and work hard. Don’t attract any attention. You should be grateful to be here.” Why do we keep quiet? Being an immigrant is a constant...

“Keep your head down, and work hard. Don’t attract any attention. You should be grateful to be here.” Why do we keep quiet? Being an immigrant is a constant...