Sum of all even integers

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Jeevanantham: Junior | Next Rank: 30 Posts; Posts: 18; Joined: Sun Apr 08, 2012 12:44 am; Thanked: 1 times

by Jeevanantham » Fri Jun 29, 2012 5:45 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) 10100
B) 20200
C) 22650
D) 40200
E) 45150

As per my understanding, I tried to solve it with the formula average = sum / number of terms. So Average = first + last / 2 = 99 +301 / 2 = 200 . Then the number of terms is Last - First / 2 + 1.
So, 301-99 / 2 + 1 = 102. Therefore Sum = 200(102) = 20400 . But the answer is 20200. Where am wrong here??

Source: Beat The GMAT — Problem Solving | Fri Jun 29, 2012 5:45 am

tutorphd: Master | Next Rank: 500 Posts; Posts: 126; Joined: Sun Jun 24, 2012 10:11 amFollowed by: 7 members; Location: Chicago, IL; Thanked: 36 times

by tutorphd » Fri Jun 29, 2012 5:58 am

Your method is correct but the first even integer in the set is 100 and the last is 300.

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nailGmat2012: Senior | Next Rank: 100 Posts; Posts: 33; Joined: Wed Jan 11, 2012 5:18 pm

by nailGmat2012 » Fri Jun 29, 2012 6:53 am

You are looking for sum of consecutive "even" integers between two odd integers.
so you are actually looking for even integers between 100 and 300 (inclusive)
Ave = Middle number = (100+300)/2 = 200
Number of even digits = (300-100)/2 + 1 = 101
sum = 200 * 101 = 20200

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16206; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Fri Jun 29, 2012 11:01 am

Jeevanantham 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) 10100
B) 20200
C) 22650
D) 40200
E) 45150

Here's a slightly different solution:

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

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16206; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Fri Jun 29, 2012 11:02 am

Alternatively, if we want to evaluate 2(50+51+52+...+149+150), we can 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 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

Cheers,
Brent

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