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Consecutive Integers

by [email protected] » Wed May 20, 2015 8:49 am

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172. PS in OG

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 - 301?

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

I don't understand the answer key in the OG Guide. Looking for a few different ways to solve.

Thanks!
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by GMATGuruNY » Wed May 20, 2015 8:54 am
For any positive integer n, the sum of first n positive integer 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
The question asks for the sum of the even integers from 100 to 300, inclusive.

Ignore the formula given. Instead, the following can be used to calculate the sum of any set of evenly spaced integers:

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)/interval + 1

The INTERVAL is the distance between one term and the next.
Since we're adding only the even integers here, the interval 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.
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by Brent@GMATPrepNow » Wed May 20, 2015 9:08 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 integers between 99 and 301 ?

(A) 10,100
(B) 20,200
(C) 22,650
(D) 40,200
(E) 45,150
Approach #1

We want 100+102+104+....298+300
This equals 2(50+51+52+...+149+150)
From here, a quick way is to evaluate this is to first recognize that there are 101 integers from 50 to 150 inclusive (150-50+1=101)

To evaluate 2(50+51+52+...+149+150) I'll add values in pairs:

....50 + 51 + 52 +...+ 149 + 150
+150+ 149+ 148+...+ 51 + 50
...200+ 200+ 200+...+ 200 + 200

How many 200's do we have in the new sum? There are 101 altogether.
101x200 = [spoiler]20,200 = B[/spoiler]

Cheers,
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by Brent@GMATPrepNow » Wed May 20, 2015 9:09 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 integers between 99 and 301 ?

(A) 10,100
(B) 20,200
(C) 22,650
(D) 40,200
(E) 45,150
Approach #2:

From my last post, we can see that we have 101 even integers from 100 to 300 inclusive.

Since the values in the set are equally spaced, the average (mean) of the 101 numbers = (first number + last number)/2 = (100 + 300)/2 = 400/2 = 200

So, we have 101 integers, whose average value is 200.
So, the sum of all 101 integers = (101)(200)
= 20,200
= B

Cheers,
Brent
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by Brent@GMATPrepNow » Wed May 20, 2015 9:09 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 integers between 99 and 301 ?

(A) 10,100
(B) 20,200
(C) 22,650
(D) 40,200
(E) 45,150
Approach #3:
Take 100+102+104+ ...+298+300 and factor out the 2 to get 2(50+51+52+...+149+150)
From here, we'll evaluate the sum 50+51+52+...+149+150, and then double it.

Important: notice that 50+51+.....149+150 = (sum of 1 to 150) - (sum of 1 to 49)

Now we use the given formula:
sum of 1 to 150 = 150(151)/2 = 11,325
sum of 1 to 49 = 49(50)/2 = 1,225

So, sum of 50 to 150 = 11,325 - 1,225 = 10,100

So, 2(50+51+52+...+149+150) = 2(10,100) = [spoiler]20,200 = B[/spoiler]

Cheers,
Brent
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by [email protected] » Thu Mar 22, 2018 12:00 pm
Hi All,

We're asked for the sum of all the EVEN integers between 99 and 301. Adding the numbers up in order would take way too much time to be practical - and the GMAT would never expect us to do the work in that way. There are actually a couple of ways to quickly add up these terms; here's one way called "bunching":

When adding up a group of numbers, the order of the numbers does NOT matter. I can bunch the numbers into consistent 'pairs'...

100 + 300 = 400
102 + 298 = 400
104 + 296 = 400
Etc.
Notice how each 'pair' of numbers sums to 400. The last 'pair' would be 198+202 = 400 and then there would be one extra term with no 'match' (the number 200). This gives us 50 'pairs' of 400 and an extra 200...(50)(400) + 200 = 20,200

Final Answer: B

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by Jeff@TargetTestPrep » Thu Jun 28, 2018 5:17 pm
[email protected] wrote:172. PS in OG

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 - 301?

A) 10,100
B) 20,200
C) 22,650
D) 40,200
E) 45,150
We are asked to find the sum of the even integers from 100 to 300 inclusive. We can use the following formula (rather than the one that has been provided):

Sum = quantity x average

The quantity is the number of even integers from 100 to 300 inclusive, which is:

(300 - 100)/2 + 1 = 101

The average of the even integers from 100 to 300, inclusive, is also the average of 100 and 300, which is:

(100 + 300)/2 = 200

Thus, the sum of the even integers from 100 to 300 inclusive is:

Sum = 101 x 200

Sum = 20,200

Answer: B

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Head of GMAT Instruction
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