It is currently 18 Nov 2017, 05:29

### 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

# Events & Promotions

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

# For any positive integer n, the sum of the first n positive

Author Message
TAGS:

### Hide Tags

Intern
Joined: 13 Feb 2012
Posts: 3

Kudos [?]: 191 [10], given: 1

For any positive integer n, the sum of the first n positive [#permalink]

### Show Tags

19 Feb 2012, 19:19
10
KUDOS
61
This post was
BOOKMARKED
00:00

Difficulty:

55% (hard)

Question Stats:

66% (01:34) correct 34% (01:51) wrong based on 2070 sessions

### HideShow timer Statistics

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

Hi,

Hoping you can explain this one to me.
[Reveal] Spoiler: OA

Kudos [?]: 191 [10], given: 1

Math Expert
Joined: 02 Sep 2009
Posts: 42249

Kudos [?]: 132567 [44], given: 12326

Re: For any positive integer n, the sum of the first n positive [#permalink]

### Show Tags

19 Feb 2012, 23:11
44
KUDOS
Expert's post
66
This post was
BOOKMARKED
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:
Even integer between 99 and 301 represent evenly spaced set (aka arithmetic progression): 100, 102, 104, ..., 300. Now, the sum of the elements in any evenly spaced set is the mean (average) multiplied by the number of terms. (Check Number Theory chapter of Math Book for more: math-number-theory-88376.html)

Average of the set: (largest+smallest)/2=(300+100)/2=200;
# of terms: (largest-smallest)/2+1=(300-100)/2+1=101 (check this: totally-basic-94862.html#p730075);

The sum = 200*101= 20,200.

Approach #2:
Using the formula of the sum of the first n positive integers: n(n+1)/2.

100+102+...+300=2(50+51+..+150). Now, the sum of the integers from 50 to 150, inclusive equals to the sum of the integers from 1 to 150, inclusive minus the sum of the integers from 1 to 49, inclusive. 2(50+51+..+150)=2*(150(150+1)/2-49(49+1)/2)=20,200.

Hope it helps.
_________________

Kudos [?]: 132567 [44], given: 12326

Manager
Joined: 03 Oct 2009
Posts: 60

Kudos [?]: 149 [12], given: 8

Re: For any positive integer n, the sum.... [#permalink]

### Show Tags

19 Feb 2012, 19:36
12
KUDOS
15
This post was
BOOKMARKED
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 = first term = 100
l = last term = 300

Total number of terms = ((300-100)/2 ) + 1 = 101

sum = n (a + l ) / 2 = 101 (100 + 300 ) / 2 = 20200

Kudos [?]: 149 [12], given: 8

Intern
Joined: 22 Apr 2012
Posts: 33

Kudos [?]: 47 [5], given: 19

GPA: 3.98
Re: What is the sum of all even integers between 99 and 301? [#permalink]

### Show Tags

03 Jun 2012, 17:53
5
KUDOS
1
This post was
BOOKMARKED
damham17 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

Would someone please explain this? I do not quite understand the official explanation. Thanks in advance.

The sum of the first n terms of an arithmetic pogression is : n/2 *(2a + (n-1)d)

where a is the first term and d is the difference between consecutive terms.

we are to find out the sum of all even integers between 99 and 301

The even numbers look like 100,102,104....300

a=100
n=101
d=2

placing these values in the formula n/2 *(2a + (n-1)d)

sum of even integers = 101/2 * (2*100 + (101-1)*2)
=20200

Choice B.

Alternative method:

The sum of first n positive integers is n(n+1)/2

The sum of first 301 integers is 301*302/2=301*151=45451
The sum of first 99 integers is 99*100/2=99*50=4950

The sum of integers between 99 and 301 = 45451-4950=40501

The above contains all odd and even integers between 99 and 301 --- odd + even = 40501 ------->eqn 1

consider the following series:

numbers in series number of odd numbers sum of odd numbers(so) sum of even numbers(se) so-se
123 2 4 2 2
12345 3 9 6 3
1234567 4 16 12 4

You can see that in the first n integers, the difference of the sum of the odd numbers - sum of even numbers = number of odd numbers in the set.

between 1 and 301 we have 151 odd numbers
between 1 and 99 we have 50 odd numbers

difference of sum of odd numbers and sum of even numbers between 99 and 301 = 151-50=101 ----------> eqn 2

consider eqn 1 and eqn 2

odd+even=40501
odd-even=101

subtract 2 from 1

2even=40400
sum of even numbers between 99 and 301=20200

Choice B

Approach 3

you know that sum of odd + even between 99 and 301 is 40501 (same from previous approach)

The answer should be roughly half of this number --- 20250
closest is B.

Would love to see a simpler more elegant solution. It will save time for us.
_________________

If I did make a valid point, would you please consider giving me a kudo. Thanks.

Kudos [?]: 47 [5], given: 19

Current Student
Joined: 12 Feb 2012
Posts: 17

Kudos [?]: 9 [4], given: 2

Schools: McCombs '17 (A)
GPA: 4
Re: For any positive integer n, the sum of the first n positive [#permalink]

### Show Tags

17 Mar 2012, 16:45
4
KUDOS
Since it is the even numbers, you can change the set to 100 to 300.

The average of the two is 200.

The amount of even integers is: ((300-100)/2)+ 1 = 101

200 * 101 = 20,200

Kudos [?]: 9 [4], given: 2

Veritas Prep GMAT Instructor
Joined: 11 Dec 2012
Posts: 313

Kudos [?]: 297 [3], given: 66

Re: For any positive integer n, the sum of the first n positive [#permalink]

### Show Tags

25 Feb 2013, 08:42
3
KUDOS
Expert's post
1
This post was
BOOKMARKED
Hi neerajeai, please note that the n(n+1)/2 shortcut formula is only applicable if the starting point is 1. Anytime you want to find the number of terms between two given numbers you should use the general formula ((first - last) / frequency) + 1. You can also multiply by the average at the end to get the sum.

In your case it is (((301-99)/2) + 1) * 200 = 102 * 200 = 20400. Answer choice C with a presumed typo in the unit digit?

Hope this helps!
-Ron
_________________

Kudos [?]: 297 [3], given: 66

Manager
Joined: 06 Jun 2012
Posts: 140

Kudos [?]: 273 [2], given: 37

For any positive integer n, the sum of the first n positive inte [#permalink]

### Show Tags

12 Feb 2013, 02:36
2
KUDOS
1
This post was
BOOKMARKED
Easier way would be using sum of all n even integers = n * (n+1)

Hence from 1-301 there are 150 even integers hence sum = 150 * 151 = 22650
From 1- 99 there are 49 even integers ( - 1 as 100 is not part of this) = 49 * 50 = 2450

So sum from 99-301 = 22650 -2450 = 20200
_________________

Please give Kudos if you like the post

Kudos [?]: 273 [2], given: 37

Senior Manager
Joined: 23 Oct 2010
Posts: 381

Kudos [?]: 403 [1], given: 73

Location: Azerbaijan
Concentration: Finance
Schools: HEC '15 (A)
GMAT 1: 690 Q47 V38
Re: For any positive integer n, the sum of the first n positive [#permalink]

### Show Tags

04 Jun 2012, 02:21
1
KUDOS
1
This post was
BOOKMARKED
the sum of 1st even numbers = n(n+1)
the sum of first even numbers till 301 is 150*(150+1) =22650
the sum of first even numbers till 99 is 49*(49+1) =2450

22650-2450=20200

this method is not perfect. I wrote it just to show one more method
_________________

Happy are those who dream dreams and are ready to pay the price to make them come true

I am still on all gmat forums. msg me if you want to ask me smth

Kudos [?]: 403 [1], given: 73

Manager
Joined: 16 Jan 2011
Posts: 111

Kudos [?]: 180 [1], given: 15

Re: For any positive integer n, the sum of the first n positive [#permalink]

### Show Tags

16 Jul 2012, 12:19
1
KUDOS
but whats the purpose of the formula in question itself?
Quote:
the sum of the first n positive integers equals n(n+1)/2

if there is no need in it?

Kudos [?]: 180 [1], given: 15

Senior Manager
Joined: 24 Aug 2009
Posts: 497

Kudos [?]: 867 [1], given: 276

Schools: Harvard, Columbia, Stern, Booth, LSB,
For any positive integer n, the sum of the first n positive [#permalink]

### Show Tags

19 Aug 2012, 13:01
1
KUDOS
This problem can be solved with an alternative formula:-
Sum of first 'n' even integers is given by - n(n+1), where n=2,4,6,......any even integer
_________________

If you like my Question/Explanation or the contribution, Kindly appreciate by pressing KUDOS.
Kudos always maximizes GMATCLUB worth
-Game Theory

If you have any question regarding my post, kindly pm me or else I won't be able to reply

Kudos [?]: 867 [1], given: 276

Manager
Joined: 30 May 2012
Posts: 220

Kudos [?]: 81 [1], given: 151

Location: United States (TX)
Concentration: Finance, Marketing
GPA: 3.3
WE: Information Technology (Consulting)
What is the sum of all even integers between 99 and 301? [#permalink]

### Show Tags

18 Nov 2014, 07:42
1
KUDOS
What is the sum of all even integers between 99 and 301?
a. 10,100
b. 20,200
c. 22,650
d. 40,200
e. 45,150

Kudos [?]: 81 [1], given: 151

Math Expert
Joined: 02 Sep 2009
Posts: 42249

Kudos [?]: 132567 [1], given: 12326

Re: For any positive integer n, the sum of the first n positive [#permalink]

### Show Tags

18 Nov 2014, 07:46
1
KUDOS
Expert's post
Blackbox wrote:
What is the sum of all even integers between 99 and 301?
a. 10,100
b. 20,200
c. 22,650
d. 40,200
e. 45,150

Merging topics. Please search before posting.
_________________

Kudos [?]: 132567 [1], given: 12326

Target Test Prep Representative
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 1684

Kudos [?]: 901 [1], given: 5

Re: For any positive integer n, the sum of the first n positive [#permalink]

### Show Tags

27 Apr 2016, 08:42
1
KUDOS
Expert's post
gwiz87 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

Hi,

Hoping you can explain this one to me.

Although a formula is provided in this problem, we can easily solve it using a different formula:

sum = (average)(quantity)

Let’s first determine the average.

In any set of numbers in an arithmetic sequence, we can determine the average using the formula:

(1st number in set + last number in set)/2

Remember, we must average the first even integer in the set and the last even integer in the set. So we have:

(100 + 300)/2 = 400/2 = 200

Next we have to determine the quantity. Once again, we include the first even integer in the set and the last even integer in the set. Thus, we are actually determining the quantity of even consecutive even integers from 100 to 300, inclusive.

Two key points to recognize:

1) Because we are determining the number of “even integers” in the set, we must divide by 2 after subtracting our quantities.

2) Because we are counting the consecutive even integers from 100 to 300, inclusive, we must “add 1” after doing the subtraction.

quantity = (300 – 100)/2 + 1

quantity = 200/2 + 1

quantity = 101

Finally, we can determine the sum.

sum = 200 x 101

sum = 20,200

Note that the reason this is easier than using the formula provided, is that the given formula would have to be applied several times since we don’t want the total of all of the first 300 numbers. We’d have to remember to subtract the sum of the first 99 and divide by two to count only the even numbers. But we’d also have to account for the fact that the first and the last number in the set are both even. So even though a formula is given, it isn’t very easy to use.

_________________

Jeffery Miller

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

Kudos [?]: 901 [1], given: 5

Director
Joined: 02 Sep 2016
Posts: 788

Kudos [?]: 43 [1], given: 274

For any positive integer n, the sum of the first n positive [#permalink]

### Show Tags

27 Jul 2017, 06:31
1
KUDOS
Bang2919 wrote:
fameatop wrote:
This problem can be solved with an alternative formula:-
Sum of first 'n' even integers is given by - n(n+1), where n=2,4,6,......any even integer

I am not able to get the right answer using this formula.
To find n:
(300-100)/2 + 1 = 101.
According to the formula mentioned it is n*(n+1) = 101 * 102. I am not understanding where i am going wrong

You are missing a very important point here.
The question has mentioned that the sum of FIRST N POSITIVE INTEGERS EQUALS n(n+1)/2.

We have to just find the sum of even nos. from 100 to 300.

100,102,104,106,.....................,300

You have found correctly that the number of terms is 101.

But this formula can only be used for first n positive integers i.e. 1,2,3,4,5................................and so on.

Here the series is not starting from 1 but from 100.

Therefore

To use this formula, we will have to first find the sum of first 150 terms and then subtract the sum of first 49 terms.

Why so?

We have to find: 100+102+104+106+......................+300

This can also be written as:
2(50+51+.................+150)

We need the sum of 50th term+51st term, and so on till 150th term.

After you subtract the sum of 1st term till 49th term from the sum of 1st term till 150th term, multiply the answer by 2.

I hope you get it.

Otherwise there are some other great methods discussed here.

One more method could be:

nth term= a+(n-1)d
300= 100+(n-1)2
n= 101

Sum= n/2 (First term+Last term)
Sum= 101/2 *(100+300) = 20200

I hope it helps.

Kudos [?]: 43 [1], given: 274

SVP
Joined: 14 Apr 2009
Posts: 2137

Kudos [?]: 1636 [0], given: 8

Location: New York, NY
Re: For any positive integer n, the sum of the first n positive [#permalink]

### Show Tags

26 Aug 2012, 10:30
1
This post was
BOOKMARKED
In this question, they give you a formula to reference: the sum of the first n positive integers is n(n+1)/2

But wait, the question is about the sum of EVEN integers, not all positive integers.

We can understand the sum of even integers between 99 and 301 by somehow rewriting in terms of the expression they gave us. They were using the terminology of the first n positive integers.

OK, well sum of even integers between 99 and 301 is the same as between 100 and 300 - inclusive.

Same as the sum of the even integers up to 300 - sum of the even integers up to 98
Same as the sum of the first 150 even integers - sum of the first 49 even integers

So how do we translate this to the expression we were given? We need to translate our "even" integer expression into "positive integers" expression with our formula.

Sum of first 150 even integers = Sum of first 150 POSITIVE integers * 2
1, 2, 3, 4, 5 ===> 2, 4, 6, 8, 10 etc

Sum of first 49 even integers = Sum of first 49 POSITIVE integers * 2

2 * (n(n+1)/2) - 2 * (n(n+1)/2)

Where n=150 in the first case, and n = 49 in the second case

= 2* (150(151)/2) - 2 * (49(50)/2)
= 150*151 - 49*50
= 50 (3*151 - 49)
= 50 (453 - 49)
= 50 (404)
= 50 (400 + 4)
= 20,000 + 200
= 20,200

Kudos [?]: 1636 [0], given: 8

Manager
Joined: 13 Feb 2012
Posts: 144

Kudos [?]: 11 [0], given: 85

Location: Italy
Concentration: General Management, Entrepreneurship
GMAT 1: 560 Q36 V34
GPA: 3.1
WE: Sales (Transportation)
Re: For any positive integer n, the sum of the first n positive [#permalink]

### Show Tags

01 Sep 2012, 06:26
Galiya wrote:
but whats the purpose of the formula in question itself?
Quote:
the sum of the first n positive integers equals n(n+1)/2

if there is no need in it?

The way I see it the formula is there just to make the question a bit harder. An evil trick, if you want.

The OG explanation is equally confusing and I personally do not think is worth the effort to try and master it, there are much "cleaner" ways to solve this problem, as many users here have shown.
_________________

"The Burnout" - My Debrief

Kudos if I helped you

Andy

Kudos [?]: 11 [0], given: 85

Non-Human User
Joined: 09 Sep 2013
Posts: 15697

Kudos [?]: 281 [0], given: 0

Re: For any positive integer n, the sum of the first n positive [#permalink]

### Show Tags

07 Nov 2014, 00:43
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Kudos [?]: 281 [0], given: 0

Manager
Joined: 30 May 2012
Posts: 220

Kudos [?]: 81 [0], given: 151

Location: United States (TX)
Concentration: Finance, Marketing
GPA: 3.3
WE: Information Technology (Consulting)
For any positive integer n, the sum of the first n positive [#permalink]

### Show Tags

18 Nov 2014, 08:13
If I were to use this formula for sum of numbers in an AP ---> $$\frac{n}{2}*(2a + (n-1)d)$$, I'd first find the number of terms in the collection using
n=$$\frac{(Last Term - First Term)}{2} - 1$$ (- 1 because I am excluding the first and last terms)

So, that would give n=$$\frac{(301-99)}{2}-1$$ = 100
Therefore,
a=99
n=100
d=2
Replacing these values in the formula n/2 *(2a + (n-1)d), we get sum of even integers =$$\frac{100}{2} * (2*99+ (100-1)*2)$$ = 19,800. Where am I wrong?

Kudos [?]: 81 [0], given: 151

Manager
Joined: 30 May 2012
Posts: 220

Kudos [?]: 81 [0], given: 151

Location: United States (TX)
Concentration: Finance, Marketing
GPA: 3.3
WE: Information Technology (Consulting)
For any positive integer n, the sum of the first n positive [#permalink]

### Show Tags

18 Nov 2014, 08:34
OK, I think I now understand where I went wrong. The question asks- Find the sum of multiples of 2. So, the first multiple of 2 in the collection begins at 100 and last one ends at 300. Therefore, your first term is 100 and last 300. The formula for n changes though:

n=$$\frac{(Last Term - First Term)}{2} + 1$$, which yields 101.

The value of a also changes because the first term is 100 and NOT 99. Sub-in those values,
$$\frac{101}{2} * (2*100+ (101-1)*2)$$ = 20,200

For reference, please see Bunuel's post here >>http://gmatclub.com/forum/how-many-multiples-of-4-are-there-between-12-and-94862.html#p730075

My question still is, how would the formula for n change if the question asks:
1.Including both the terms
2. Including at least one term.

Kudos [?]: 81 [0], given: 151

Senior Manager
Joined: 13 Jun 2013
Posts: 278

Kudos [?]: 479 [0], given: 13

For any positive integer n, the sum of the first n positive [#permalink]

### Show Tags

18 Nov 2014, 08:55
Blackbox wrote:
If I were to use this formula for sum of numbers in an AP ---> $$\frac{n}{2}*(2a + (n-1)d)$$, I'd first find the number of terms in the collection using
n=$$\frac{(Last Term - First Term)}{2} - 1$$ (- 1 because I am excluding the first and last terms)

So, that would give n=$$\frac{(301-99)}{2}-1$$ = 100
Therefore,
a=99
n=100
d=2
Replacing these values in the formula n/2 *(2a + (n-1)d), we get sum of even integers =$$\frac{100}{2} * (2*99+ (100-1)*2)$$ = 19,800. Where am I wrong?

the formula, you are using for calculating the total number of terms is wrong. because you shouldn't put '-1' at the end of expression. why ??
consider this. for any arithmetic sequence, nth term is given as tn= a+(n-1)d. now here, if we consider a=99 and tn+1=301. and d=1, that is all terms are consecutive, then we have n-1=301-99, n=203. i.e we have total of 203 terms in the expression. if i remove 99 from the sequence. ( which you were trying to do in a wrong way.). then sequence will begin from 100 and ends at 301. i.e we have total of 202 terms. now of these terms. half are even, and half are odd. i.e. we have 101 even terms and 101 odd terms

now put the same thing in your formula of n/2 *(2a + (n-1)d), n=101 and a=100 (why ?? because we have already discarded 99) and d=2
101/2(200+200)=20,200

Kudos [?]: 479 [0], given: 13

For any positive integer n, the sum of the first n positive   [#permalink] 18 Nov 2014, 08:55

Go to page    1   2    Next  [ 30 posts ]

Display posts from previous: Sort by