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

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Vincen: Legendary Member; Posts: 2898; Joined: Thu Sep 07, 2017 2:49 pmFollowed by: 5 members; Thanked: 6 times

by Vincen » Tue Jan 19, 2021 9:47 am

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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: Beat The GMAT — Problem Solving | Tue Jan 19, 2021 9:47 am

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8246; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

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

by Scott@TargetTestPrep » Wed Jan 27, 2021 5:26 am

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

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