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Consecutive Integers

This topic has 4 expert replies and 0 member replies

Consecutive Integers

Post Wed May 20, 2015 8:49 am
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!
B

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GMAT/MBA Expert

Post Wed May 20, 2015 9:09 am
Quote:
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) = 20,200 = B

Cheers,
Brent

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GMAT/MBA Expert

Post Wed May 20, 2015 9:09 am
Quote:
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

_________________
Brent Hanneson – Founder of GMATPrepNow.com
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GMAT/MBA Expert

Post Wed May 20, 2015 8:54 am
Quote:
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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GMAT/MBA Expert

Post Wed May 20, 2015 9:08 am
Quote:
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 = 20,200 = B

Cheers,
Brent

_________________
Brent Hanneson – Founder of GMATPrepNow.com
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