GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 17 Nov 2019, 01:29 ### 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

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # The sum of the first 50 positive even integers is 2550. What is the

Author Message
TAGS:

### Hide Tags

Manager  Joined: 20 Aug 2009
Posts: 92
The sum of the first 50 positive even integers is 2550. What is the  [#permalink]

### Show Tags

2
20 00:00

Difficulty:   35% (medium)

Question Stats: 74% (01:54) correct 26% (02:00) wrong based on 654 sessions

### HideShow timer Statistics

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

A. 5,100
B. 7,550
C. 10,100
D. 15,500
E. 20,100

Originally posted by CasperMonday on 03 Sep 2009, 06:35.
Last edited by Bunuel on 11 Oct 2019, 05:29, edited 4 times in total.
Renamed the topic and edited the question.
Math Expert V
Joined: 02 Sep 2009
Posts: 59087
The sum of the first 50 positive even integers is 2550. What is the  [#permalink]

### Show Tags

6
6
enigma123 wrote:
The sum of the first 50 positive even integers is 2550. What is the sum of the even integers from 102 to 200 inclusive?

A. 5,100
B. 7,550
C. 10,100
D. 15,500
E. 20,100

# of even terms (multiples of 2) in the range from 102 to 200, inclusive is $$\frac{(last - first)}{2} + 1=\frac{(200-102)}{2}+1=50$$;

$$Mean = \frac{(last+first)}{2}$$;

$$The \ sum = (mean)*(number \ of \ terms) = \frac{(200+102)}{2}*50 = 7,550$$.

_________________
Manager  Joined: 18 Aug 2009
Posts: 238
Re: The sum of the first 50 positive even integers is 2550. What is the  [#permalink]

### Show Tags

17
4
CasperMonday wrote:
The sum of the first 50 positive even integers is 2550. What is the sum of even integers from 102 to 200 inclusive?
a. 5100
b. 7550
c. 10100
d. 15500
e. 20100

OA

I had derived a formula for calculating the summation of even/odd numbers in a series, when I was in high school.

Sum of EVEN numbers in a series from 1 -> n (where n is the LAST EVEN number in a series) = n*(n+2)/4
Sum of ODD numbers in a series from 1 -> n (where n is the LAST ODD number in a series) = (n+1)*(n+1)/4

This works perfect for any series starting from 1. Using the above formula we can solve the problem:

200*202/4 - 100*102/4 = 7550

##### General Discussion
Manager  Joined: 20 Aug 2009
Posts: 92
Re: The sum of the first 50 positive even integers is 2550. What is the  [#permalink]

### Show Tags

9
5
My solution is:

First 50 even integers:
2
4
6
8
<...>

Integers from 102 to 200
102
104
106
108
<...>

We notice that each integer from the second set is 100 more than the respective integer in the first set. Since we have 50 even integers from 102 to 200, then:
2550+(100*50)=7550.

Are there any other ways to solve the prob? Thanks!

_________________________________
Please kudos me if you find my post useful
Intern  Joined: 24 Aug 2009
Posts: 29
Re: The sum of the first 50 positive even integers is 2550. What is the  [#permalink]

### Show Tags

5
5
CasperMonday wrote:
The sum of the first 50 positive even integers is 2550. What is the sum of even integers from 102 to 200 inclusive?
a. 5100
b. 7550
c. 10100
d. 15500
e. 20100

OA

Sum of even integers from 102-200(both included)

No of terms = (200-102)/2 + 1 = 50 (+1 because both are included)

If you observe, this is AP (Arithmetic progression): 102,104,106,......,200
Sum of n terms in AP Formula, Sum = n/2[2a + (n-1)d]
where,
n = no of terms = 50 in this case.
a = starting number = 102
d = difference between AP = 2

So, Sum = 50/2[2*102 + (50-1)2]

Solving this, we can get the answer.

This is one more solution for this kind of problems.
Intern  Joined: 30 Jun 2009
Posts: 36
Re: The sum of the first 50 positive even integers is 2550. What is the  [#permalink]

### Show Tags

2
1
Sum of even integers from 102-200(both included)

No of terms = (200-102)/2 + 1 = 50 (+1 because both are included)

If you observe, this is AP (Arithmetic progression): 102,104,106,......,200
Sum of n terms in AP Formula, Sum = n/2[2a + (n-1)d]
where,
n = no of terms = 50 in this case.
a = starting number = 102
d = difference between AP = 2

So, Sum = 50/2[2*102 + (50-1)2]

Solving this, we can get the answer.

This is one more solution for this kind of problems.

Saurabhricha,
what would be the formula for odd nbrs?
Manager  Joined: 28 Jul 2011
Posts: 155
Re: The sum of the first 50 positive even integers is 2550. What is the  [#permalink]

### Show Tags

3
Sum = Avg * number of items

Avg = (200 + 102) / 2 = 302 / 2 = 151

No. of items = 200-102 = 98/2 (Only the even numbers so divide by 2) = 49 + 1 (add 1 the last item) = 50

Sum = 151 * 50 = 7550

ANS : B
Intern  Joined: 19 Dec 2012
Posts: 18
Re: The sum of the first 50 positive even integers is 2550. What is the  [#permalink]

### Show Tags

6
2
Even easier method:

First 50 even integers are 2, 4, 6, ... , 100. Sum is stated as 2550.
Sum of even integers between 102 and 200 are 100 more than each integer in the previous sum (102 is 100 more than 2, etc.).

So...when this happens 50 times, it's 50*100 = 5000, 5000 + 2550 = 7550.

No need for formula, just simple number pattern here.

---
My GMAT Experience: 670-to-760-q49-v44-ir-8-awa-6-0-no-stone-unturned-for-8-months-200051.html

Originally posted by rf3d3r3r on 28 Jan 2013, 11:19.
Last edited by rf3d3r3r on 16 Jun 2015, 22:01, edited 1 time in total.
Intern  Status: preparing for the GMAT
Joined: 16 Jul 2013
Posts: 28
Concentration: Technology, Entrepreneurship
GMAT Date: 10-15-2013
GPA: 3.53
Re: The sum of the first 50 positive even integers is 2550. What is the  [#permalink]

### Show Tags

I am totally confused. in the math book, the formula for the first n even number is n(n+1)

and here there is another formula used, mean*# of terms

_________________
لا الله الا الله, محمد رسول الله

You never fail until you stop trying ,,,
Math Expert V
Joined: 02 Sep 2009
Posts: 59087
Re: The sum of the first 50 positive even integers is 2550. What is the  [#permalink]

### Show Tags

1
Aldossari wrote:
I am totally confused. in the math book, the formula for the first n even number is n(n+1)

and here there is another formula used, mean*# of terms

The sum of n first positive even numbers is n(n+1) --> the sum of first 50 positive even numbers is 50*51 = 2,550
The sum of evenly spaced set is (mean)*(# of terms) --> the sum of first 50 positive even numbers is (2+100)/2*50 = 51*50 = 2,550.
_________________
SVP  Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1729
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: The sum of the first 50 positive even integers is 2550. What is the  [#permalink]

### Show Tags

3
2 + 4 + 6 + 8 ................... + 100 = 2550 ................. (1)

102 + 104 + 106 + 108 ............ + 200 = ?? ................... (2)

Just note that we have to add 100 to each term of series (1) to get the corresponding terms in series (2)

(2+100) + (4+100) + (6+100) + (8+100) ................ + (100+100)

This 100 ha to be added 50 times, so result of series (2) would be

2550 + 50*100 = 7550

_________________
Kindly press "+1 Kudos" to appreciate GMAT Club Legend  V
Joined: 12 Sep 2015
Posts: 4064
Re: The sum of the first 50 positive even integers is 2550. What is the  [#permalink]

### Show Tags

Top Contributor
enigma123 wrote:
The sum of the first 50 positive even integers is 2550. What is the sum of the even integers from 102 to 200 inclusive?

A. 5,100
B. 7,550
C. 10,100
D. 15,500
E. 20,100

The sum of the first 50 positive even integers is 2550.
In other words, 2 + 4 + 6 + 8 + ...+ 98 + 100 = 2550

We want the sum: 102 + 104 + 106 + . . . 198 + 200
IMPORTANT: Notice that each term in this sum is 100 GREATER than each term in the first sum.
In other words, 102 + 104 + 106 + . . . 198 + 200 is the SAME AS...
(100 + 2) + (100 + 4) + (100 + 6) + ... + (100 + 98) + (100 + 100)
We can rearrange these terms to get: (100 + 100 + ... + 100 + 100) + (2 + 4 + 6 + 8 + ...+ 98 + 100)

IMPORTANT: There are 50 100's in the red sum, and we're told that the blue sum = 2550

So, our sum = 50(100) + 2550
= 5000 + 2550
= 7550

RELATED VIDEO

_________________
Senior Manager  G
Joined: 04 Aug 2010
Posts: 492
Schools: Dartmouth College
Re: The sum of the first 50 positive even integers is 2550. What is the  [#permalink]

### Show Tags

enigma123 wrote:
The sum of the first 50 positive even integers is 2550. What is the sum of the even integers from 102 to 200 inclusive?

A. 5,100
B. 7,550
C. 10,100
D. 15,500
E. 20,100

For any EVENLY SPACED SET:
Count = $$\frac{(biggest - smallest)}{(increment)}$$ + 1
Average = median = $$\frac{(biggest + smallest)}{2}$$
Sum = (count)(average)
The INCREMENT is the difference between successive values.

Even numbers between 102 and 200, inclusive:
Here, the integers are EVEN, so the increment = 2.
Count = $$\frac{(200-102)}{2}$$ + 1 = 50.
Average = $$\frac{(200+102)}{2}$$ = 151.
Sum = (50)(151) = 7750.

Alternate approach:

102, 104, 106......196, 198, 200.

The median is halfway between 102 and 200:
$$\frac{(102+200)}{2}$$= 151.
Of the 100 integers between 101 and 200, inclusive, half are odd, half are even.
Thus, in the set above, there are 25 even integers to the LEFT of 151 and 25 even integers to the RIGHT of 151, for a total of 50 even integers.

Add successive PAIRS of integers, working from the OUTSIDE IN:
102+200 = 302.
104+198 = 302.
106+196 = 302.

Notice that the sum of each pair = 302.
Since there will be 25 of these pairs, the following sum is yielded:
25*302 = 7550.

_________________
GMAT and GRE Tutor
Over 1800 followers
GMATGuruNY@gmail.com
New York, NY
If you find one of my posts helpful, please take a moment to click on the "Kudos" icon.
Available for tutoring in NYC and long-distance.
VP  D
Joined: 09 Mar 2016
Posts: 1229
Re: The sum of the first 50 positive even integers is 2550. What is the  [#permalink]

### Show Tags

enigma123 wrote:
The sum of the first 50 positive even integers is 2550. What is the sum of the even integers from 102 to 200 inclusive?

A. 5,100
B. 7,550
C. 10,100
D. 15,500
E. 20,100

SUM OF N TERMS = $$\frac{n}{2} (2a+(n-1)d)$$

$$a$$ = 102 the first term
$$d$$ = 2 distance
$$n$$ = 50 how many terms to add up

$$\frac{50}{2} (2 (102)+(50-1)2)$$ = 7550

another approach Sum = average x number of terms

Average = $$\frac{(200 + 102)}{2} = 151$$

Number of terms $$\frac{(200-102)}{2} + 1 = 50$$

Sum of Even integers 151 *50 = 7,550 VP  D
Joined: 14 Feb 2017
Posts: 1281
Location: Australia
Concentration: Technology, Strategy
GMAT 1: 560 Q41 V26 GMAT 2: 550 Q43 V23 GMAT 3: 650 Q47 V33 GMAT 4: 650 Q44 V36 GMAT 5: 650 Q48 V31 GPA: 3
WE: Management Consulting (Consulting)
Re: The sum of the first 50 positive even integers is 2550. What is the  [#permalink]

### Show Tags

[[(200-102)/2 ] +1]*(200+102)/2
_________________
Goal: Q49, V41

+1 Kudos if I have helped you Re: The sum of the first 50 positive even integers is 2550. What is the   [#permalink] 16 Oct 2019, 14:53
Display posts from previous: Sort by

# The sum of the first 50 positive even integers is 2550. What is the  