For any positive integer \(n,\) the sum of the first \(n\) positive integers equals \(\dfrac{n(n+1)}2.\) What is the sum

This topic has expert replies
Legendary Member
Posts: 2898
Joined: Thu Sep 07, 2017 2:49 pm
Thanked: 6 times
Followed by:5 members

Timer

00:00

Your Answer

A

B

C

D

E

Global Stats

For any positive integer \(n,\) the sum of the first \(n\) positive integers equals \(\dfrac{n(n+1)}2.\) 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

Answer: B

Source: Official Guide
Source: — Problem Solving |

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 8086
Joined: Sat Apr 25, 2015 10:56 am
Location: Los Angeles, CA
Thanked: 43 times
Followed by:29 members
Vincen wrote:
Tue Jan 19, 2021 9:47 am
For any positive integer \(n,\) the sum of the first \(n\) positive integers equals \(\dfrac{n(n+1)}2.\) 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

Answer: B

Source: Official Guide
Solution:

The phrase “between 99 and 301” does not include the endpoints. The sum of the even integers from 100 to 300 is:

[(100 + 300)/2] * [(300 - 100)/2 + 1]

= 200 * 101

= 20,200

[Note: The first factor is the average of the even integers and the second factor is the number of even integers from 100 to 300.]

Answer: B

Scott Woodbury-Stewart
Founder and CEO
[email protected]

Image

See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews

ImageImage