It is currently 24 Jun 2017, 10:34

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# ps:sum

Author Message
Senior Manager
Joined: 22 Jun 2005
Posts: 363
Location: London

### Show Tags

26 May 2006, 14:08
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 100% (01:04) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

The sum of the first 50 positive even integers is 2550. What is the sum of the even integers from 102 to 200, inclusive?

1. 5100
2. 7550
3. 10100
4. 15500
5. 20100

Senior Manager
Joined: 09 Mar 2006
Posts: 444

### Show Tags

26 May 2006, 14:25
102 + 104 + ... + 200 = 100 * 50 + 2 * ( 1 + 2 + ... + 50 ) = 5000 + 2 * 2550 = 10100
Director
Joined: 13 Nov 2003
Posts: 789
Location: BULGARIA

### Show Tags

26 May 2006, 22:29
Here is my piece.
From 102 to 200 inclusive, there are 99 integers, 50 even and 49 odd. The sum of the first even and last even is
102+200=302
104+198=302
106+196=302 and so on you have got 25 pairs with sum 302 or 7550
Senior Manager
Joined: 09 Mar 2006
Posts: 444

### Show Tags

27 May 2006, 00:44
BG wrote:
Here is my piece.
From 102 to 200 inclusive, there are 99 integers, 50 even and 49 odd. The sum of the first even and last even is
102+200=302
104+198=302
106+196=302 and so on you have got 25 pairs with sum 302 or 7550

You are right, I got the question wrong since 2550 is not the sum of
first 50 integers, but that of first 50 even integers ( 2+4+...+100 )

Well done!
Manager
Joined: 05 Jan 2006
Posts: 66

### Show Tags

27 May 2006, 03:39
I got B. Just add 100 to each of the previous 50 even integers. Since there are 50 terms, you add 5000 to the previous total, which makes B the logical answer.
27 May 2006, 03:39
Display posts from previous: Sort by