BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

Sum of even integer

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Junior | Next Rank: 30 Posts
Posts: 11
Joined: Sun Nov 21, 2010 8:07 pm
Thanked: 2 times
by sodha.rakesh » Tue Jan 18, 2011 9:57 am
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 integer between 99 and 301?

A. 10,100
B. 20,200
C. 22,650
D. 40,200
E. 45,150
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members
Location: New York, NY
Thanked: 13060 times
GMAT Score:790

by GMATGuruNY » Tue Jan 18, 2011 10:15 am
sodha.rakesh 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 integer between 99 and 301?

A. 10,100
B. 20,200
C. 22,650
D. 40,200
E. 45,150
We need to count the even integers from 100 to 300, inclusive.

Ignore the formula given. Instead, use the following:

Sum = (number of integers) * (average of biggest and smallest)

To count the number of evenly spaced integers in a set:

Number of integers = (Biggest - Smallest)/(distance between each successive pair) + 1

Since we're adding only the even integers, the distance between each successive pair is 2.
Thus, the number of even integers from 100 to 300 = (300-100)/2 + 1 = 101.
Average of biggest and smallest = (300+100)/2 = 200.
Sum = (number of integers) * (average of biggest and smallest) = 101*200 = 20,200.

The correct answer is B.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 233
Joined: Mon Jun 09, 2008 1:30 am
Thanked: 5 times

by blaster » Tue Jan 18, 2011 11:00 am
the simple logic of this question is you should find all the numers sum from 1...301, also find 1..99 then the first result minus the second. the number you will get will be around 40000 . but in question we have additional restriction. the even numbers. as in a number line we have one odd and one even you need just divide the result to 2.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3835
Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members
Location: Milpitas, CA
Thanked: 1854 times
GMAT Score:770

by Anurag@Gurome » Tue Jan 18, 2011 11:30 am
blaster wrote:the simple logic of this question is you should find all the numers sum from 1...301, also find 1..99 then the first result minus the second. the number you will get will be around 40000 . but in question we have additional restriction. the even numbers. as in a number line we have one odd and one even you need just divide the result to 2.
Not so simple!
Just consider the integers from 1 to 10.
Sum of all integers = 55
Sum of odd integers = 25 ≠ (55/2)
Sum of even integers = 30 ≠ (55/2)

In fact in this case following your method will result a sum equal to 20,250.5, which is not correct.

If you have to use the given formula, then it can be done as follows,
  • .. Sum of all the even integer between 99 and 301
    = (100 + 102 + 104 + ... + 300)
    = 2*(50 + 51 + 52 + ... + 150)
    = 2*[(1 + 2 + 3 + ... + 150) - (1 + 2 + 3 + ... + 49)]
    = 2*[(150*151)/2 - (49*50)/2]
    = [(150*151) - (49*50)]
    = 50*(3*151 - 49)
    = 50*404
    = 20,200
The solution provided by Mitch is more efficient and time saving.
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Newbie | Next Rank: 10 Posts
Posts: 1
Joined: Thu Jan 13, 2011 5:55 am

by hovhannest » Tue Jan 18, 2011 1:52 pm
Arithmetic progression... sum = (a1+an)*n/2, for our case sum = (100+300)*n/2, where n=101 (note please that between [100 - 300] 100 odd and 101 even numbers).

Thus, 200*101=20,200
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1