Kevdog2834 wrote:Official Guide: 12th Edition
Problem Solving
#157

Can someone please help explain this problem to me. Some of the background information on the topic would be great. The explanation in the book uses Calculus and there is definitely an easier way.

How does it work with:
[Num Terms * Avg(High + Low term)] / 2

After that the number is suppose to be doubled?

dude post problem

When trying to determine the sum of a set of evenly spaced integers, the key formula to keep in mind is: Average of items in the set * Number of items in the set = Sum of all the items in the set.

To figure out the average of a set of evenly-spaced integers, add up the first and last integer and divide by 2. In this case, we can disregard the first and last integers of the set (99 and 301) since the question is asking us about the sum of the EVEN integers. So, we do (300 + 100)/2 = 200.

To get the number of items in a set of evenly-spaced integers, use the formula: [(Greatest Integer - Smallest Integer)/Increment] + 1. In this case, the greatest integer is 300 and the smallest integer is 100 (since we're dealing with only even integers). The increment here is 2 since even integers are separated by 2. Thus, there are [(300 - 100)/2] + 1 integers in this set: 101.

To get the sum of these integers, we multiply the average (200) x number of integers (101) to arrive at 20,200.

The correct answer is B.

It's worth noting that this is the rare GMAT question in which the information provided in the question is not necessary to answer the problem.

Kevdog2834 wrote:Official Guide: 12th Edition
Problem Solving
#157

Can someone please help explain this problem to me. Some of the background information on the topic would be great. The explanation in the book uses Calculus and there is definitely an easier way.

How does it work with:
[Num Terms * Avg(High + Low term)] / 2

After that the number is suppose to be doubled?

amising6 wrote:
dude post problem

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

You have 1+2+3+........ on til n terms
There is formula to resolve this exp above

n(n+1) / 2

Sum of the even integers from 100 to 300 = between 99 to 301
You have to find the first 150 even integers
then
substract the sum of the first 49 even integers.

Just apply the formula above
=2(150(150+1) /2) - 2(49(49+1) / 2)
I just applied the formula then the answer is 20200

dude [Num Terms * Avg(High + Low term)] / 2
it is just arythmetic progression that you ve seen in high school I'm sure!
T =1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9
T =9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1
2 T =10 + 10 + 10 + 10 + 10 + 10 + 10 + 10
2 T =9 x 10
T =9 x 5

You can see that T =9 x ( 9 + 1 ) / 2

1 + 2 + 3 + ... + n = n ( n + 1 ) / 2

That's it !!!

OG 12, #157

A 30-second hack, 2 detailed ways to solve, and 3 take-Away lessons are attached. If you cannot see the attachment, read it here.

-Patrick

Great file Patrick. That is a huge help.

Does there happen to be a place full of files organized in that arrangement? It was a very easy way to learn.

Hi Kev,

You're welcome. The file is an excerpt of the GMATFix OG Companion. You can find the book by clicking the book title at the top of the file. The end of the book has a reference section that organizes all the OG questions by difficulty and by topic.

I'm glad you found the explanation helpful

-Patrick

Patrick,

I believe there is a typo on solution B.




Notice you say that the highest number is 200. I believe you meant to say it was 300. Obviously you plugged in the correct number to the formula.