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
OAB
Can you please check my solution and let me know where i am getting wrong.
# of terms = (300-100)/2 + 1 = 101
Sum = n(n+1)/2
101(102)/2
101X51
Doing this unit digit will be 1, which is not in the options.
Please explain.
What is the sum of all the even integers between 99 and 301?
This topic has expert replies
GMAT/MBA Expert
- Brent@GMATPrepNow
- GMAT Instructor
- Posts: 16207
- Joined: Mon Dec 08, 2008 6:26 pm
- Location: Vancouver, BC
- Thanked: 5254 times
- Followed by:1268 members
- GMAT Score:770
You are correct to say that there are 101 terms in the sequence.kamalj wrote: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
OAB
Can you please check my solution and let me know where i am getting wrong.
# of terms = (300-100)/2 + 1 = 101
Sum = n(n+1)/2
101(102)/2
101X51
Doing this unit digit will be 1, which is not in the options.
Please explain.
However, the formula you used (sum = n(n+1)/2) does not necessarily apply here.
That formula tells us the sum of the first n CONSECUTIVE integers.
That is, 1 + 2 + 3 + 4 + 5 + . . . + n = n(n+1)/2
So, for example, 1 + 2 + 3 + 4 + 5 + . . . + 101 = 101(101+1)/2
BUT, this is not what you are asked to find.
You are asked to find the sum of 100 + 102 + 104 + 106 + . . . . 298 + 300 (not the sum 1 + 2 + 3 + 4 + 5 + . . . + 101)
Does that help?
Cheers,
Brent
GMAT/MBA Expert
- Brent@GMATPrepNow
- GMAT Instructor
- Posts: 16207
- Joined: Mon Dec 08, 2008 6:26 pm
- Location: Vancouver, BC
- Thanked: 5254 times
- Followed by:1268 members
- GMAT Score:770
Approach #1For 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
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,
Brent
GMAT/MBA Expert
- Brent@GMATPrepNow
- GMAT Instructor
- Posts: 16207
- Joined: Mon Dec 08, 2008 6:26 pm
- Location: Vancouver, BC
- Thanked: 5254 times
- Followed by:1268 members
- GMAT Score:770
Approach #2: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
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
GMAT/MBA Expert
- Brent@GMATPrepNow
- GMAT Instructor
- Posts: 16207
- Joined: Mon Dec 08, 2008 6:26 pm
- Location: Vancouver, BC
- Thanked: 5254 times
- Followed by:1268 members
- GMAT Score:770
Approach #3: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
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
GMAT/MBA Expert
- Scott@TargetTestPrep
- GMAT Instructor
- Posts: 7223
- Joined: Sat Apr 25, 2015 10:56 am
- Location: Los Angeles, CA
- Thanked: 43 times
- Followed by:29 members
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):kamalj wrote: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
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
Scott Woodbury-Stewart
Founder and CEO
[email protected]
See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews