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

It is currently 17 Nov 2018, 03:35

Close

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
Your Progress

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.

Close

Request Expert Reply

Confirm Cancel
Events & Promotions in November
PrevNext
SuMoTuWeThFrSa
28293031123
45678910
11121314151617
18192021222324
2526272829301
Open Detailed Calendar
  • FREE Quant Workshop by e-GMAT!

     November 18, 2018

     November 18, 2018

     07:00 AM PST

     09:00 AM PST

    Get personalized insights on how to achieve your Target Quant Score. November 18th, 7 AM PST
  • How to QUICKLY Solve GMAT Questions - GMAT Club Chat

     November 20, 2018

     November 20, 2018

     09:00 AM PST

     10:00 AM PST

    The reward for signing up with the registration form and attending the chat is: 6 free examPAL quizzes to practice your new skills after the chat.

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

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Director
Director
avatar
G
Joined: 02 Sep 2016
Posts: 688
Premium Member
Re: For any positive integer n, the sum of the first n positive integers  [#permalink]

Show Tags

New post 04 May 2017, 23:28
karishma Bunuel

The formula to find the sum of FIRST n nos. = n(n+1)/2

Set 1: 1,2,3,4...........10
n= 10 (no. of terms)

Sum (using the above formula)= 10*11/2= 55


Set 2: 11,12,13,14,...............20
n( again)= 10

Sum (using the above formula)= 10*11/2= 55

Which is wrong because the sum of set 2 is 155 and not 55.

My query is: This formula can be used only to find the sum of n nos. starting from 1 ?

That is why in the solution given by Bunuel:
As we have to find the sum of nos. from 50 to 150
Step 1: Sum of nos. from 1 to 150 (Sum 1)
Step 2: Sum of nos. from 1 to 49 (Sum 2)
Step 3: Subtract Sum 2 from sum 1 to get sum of nos. from 50 to 150

1,2,3,4........................49, 50, 51,...............150
_________________

Help me make my explanation better by providing a logical feedback.

If you liked the post, HIT KUDOS !!

Don't quit.............Do it.

Director
Director
avatar
G
Joined: 02 Sep 2016
Posts: 688
Premium Member
Re: For any positive integer n, the sum of the first n positive integers  [#permalink]

Show Tags

New post 04 May 2017, 23:32
Cez005 wrote:
Bunuel 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

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: http://gmatclub.com/forum/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: http://gmatclub.com/forum/totally-basic ... ml#p730075);

The sum = 200*101= 20,200.

Answer: B.

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.

Answer: B.

Hope it helps.


Hi Bunuel,

The first explanation is fantastic; however, I'm still a bit confused about the second approach. Can you clarify why finding the sum of the integers from 50 to 150 is sufficient?


Finding the sum of integers from 50 to 150 is sufficient because:

We have to find sum of: 100+102+104+...........300 (even nos.)

What is common in all even nos. ?
They are all multiple of 2 i.e we can take 2 common.

That is what Bunuel did in this question.

Sum= 100+102+104+............+300
Sum= 2(50+51+...............+150)

It's the same thing.
2*50 = 100 (first term in the series)
.
.
.
.
.
2*150= 300 (last term in the series)
Intern
Intern
avatar
B
Joined: 23 Aug 2015
Posts: 24
Location: India
Concentration: General Management, Human Resources
GMAT 1: 610 Q46 V29
GPA: 3
WE: Consulting (Human Resources)
Re: For any positive integer n, the sum of the first n positive integers  [#permalink]

Show Tags

New post 22 Jun 2017, 23:02
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 :(
Director
Director
avatar
G
Joined: 02 Sep 2016
Posts: 688
Premium Member
Re: For any positive integer n, the sum of the first n positive integers  [#permalink]

Show Tags

New post 27 Jul 2017, 05:31
1
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.
Study Buddy Forum Moderator
User avatar
D
Joined: 04 Sep 2016
Posts: 1251
Location: India
WE: Engineering (Other)
Premium Member CAT Tests
Re: For any positive integer n, the sum of the first n positive integers  [#permalink]

Show Tags

New post 28 Dec 2017, 18:06
Bunuel, niks18


Quote:
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.




This goes in to complex multiplication of a 2 digit with 3 digit, (even taking out common did not help)
Any easier way for such calculations?
_________________

It's the journey that brings us happiness not the destination.

Manager
Manager
avatar
B
Joined: 23 Oct 2017
Posts: 64
Re: For any positive integer n, the sum of the first n positive integers  [#permalink]

Show Tags

New post 28 Dec 2017, 18:19
It is clear that to calculate the sum of even integers from 99 to 301:
a= 100; d=2;
To calculate the number of terms we know the last term = 300
100 + (n-1)*2 = 300 => n= 101

sum =(n/2)*(2a + (n-1)*d)
=(101/2)*(200+100*3) = 20200
PS Forum Moderator
avatar
D
Joined: 25 Feb 2013
Posts: 1217
Location: India
GPA: 3.82
GMAT ToolKit User Premium Member Reviews Badge
Re: For any positive integer n, the sum of the first n positive integers  [#permalink]

Show Tags

New post 29 Dec 2017, 10:51
adkikani wrote:
Bunuel, niks18


Quote:
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.




This goes in to complex multiplication of a 2 digit with 3 digit, (even taking out common did not help)
Any easier way for such calculations?


Hi adkikani

you can use AP formula to find the number of even terms bewteen 99 & 301 and then use the sum formula for an AP series to calcualte the final sum

\(T_n=a_1+(n-1)d\), here \(T_n=300, a_1=100, d=2\), find \(n\)

then Sum, \(S_n=\frac{n(T_n+a_1)}{2}\).

the calculation should be simple.
Study Buddy Forum Moderator
User avatar
D
Joined: 04 Sep 2016
Posts: 1251
Location: India
WE: Engineering (Other)
Premium Member CAT Tests
Re: For any positive integer n, the sum of the first n positive integers  [#permalink]

Show Tags

New post 31 Dec 2017, 16:29
niks18

I was confused if the second formula was applied only when my sequence is starting from 1.
_________________

It's the journey that brings us happiness not the destination.

PS Forum Moderator
avatar
D
Joined: 25 Feb 2013
Posts: 1217
Location: India
GPA: 3.82
GMAT ToolKit User Premium Member Reviews Badge
Re: For any positive integer n, the sum of the first n positive integers  [#permalink]

Show Tags

New post 31 Dec 2017, 20:19
adkikani wrote:
niks18

I was confused if the second formula was applied only when my sequence is starting from 1.


Hi adkikani

The formula is a simple derivation of AP series that starts from 1 and is consecutive

So the Sum of AP series is \(S_n=\frac{n[2a+(n-1)d]}{2}=\frac{n}{2}\)[First term + Last Term]

if the sequence starts from 1 and is consecutive so your First term=1 and Last Term=number of terms in the series (because sequence is consecutive)

so \(S_n=\frac{n(1+n)}{2}\)

Let's take a simple example. if I have to find sum of numbers from 6 to 10, then I can do that by first finding sum of all Numbers from 1 to 10 and then from this sum subtract sum of numbers from 1 to 5.

In the number line this can be represented as

1...2...3...4...5...6...7...8...9...10. Here First term =1 and last term=10=number of terms in the sequence

As you can clearly see to find sum of numbers from 6 to 10, we simply need to remove sum of numbers from 1 to 5 out of the total sum i.e from 1 to 10.

Now in this question we need to find sum of number from 50 to 150 so it will be Total sum of numbers from 1 to 150 less sum of numbers from 1 to 49
CEO
CEO
User avatar
D
Joined: 11 Sep 2015
Posts: 3122
Location: Canada
Re: For any positive integer n, the sum of the first n positive integers  [#permalink]

Show Tags

New post 19 Apr 2018, 12:46
Top Contributor
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


Here's one approach.

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), let's 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.
101 x 200 = 20,200

Answer: B

Cheers,
Brent
_________________

Test confidently with gmatprepnow.com
Image

CEO
CEO
User avatar
D
Joined: 11 Sep 2015
Posts: 3122
Location: Canada
Re: For any positive integer n, the sum of the first n positive integers  [#permalink]

Show Tags

New post 19 Apr 2018, 12:47
Top Contributor
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


Approach #2:

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
_________________

Test confidently with gmatprepnow.com
Image

CEO
CEO
User avatar
D
Joined: 11 Sep 2015
Posts: 3122
Location: Canada
Re: For any positive integer n, the sum of the first n positive integers  [#permalink]

Show Tags

New post 19 Apr 2018, 12:48
Top Contributor
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


Approach #3:
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) = 20,200

Answer: B

Cheers,
Brent
_________________

Test confidently with gmatprepnow.com
Image

Intern
Intern
User avatar
B
Joined: 14 Feb 2016
Posts: 45
Re: For any positive integer n, the sum of the first n positive integers  [#permalink]

Show Tags

New post 20 Apr 2018, 00:57
Step 1: The formula is unnecesary and in my opinion is thrown there to put us offguard. We can solve this using the following two formulae:
Step 2: Find amount of integeres between 99 and 301. (First-Last/Increment+1) = 101
Step 3: Find the average of the set of numbers between 99 and 301. Since this is an evenly spaced set (constant increments of 2) We can find the mean of this number set by taking the average of the first and last terms. 99+301 = 400 / 2 = 200.
Step 4: Multiply the mean * number of integers in the set = 200*101 = 20,200.
Answer choice B.
_________________

Went from a score from 320 to 600. Still climbing! Onto the top 1%.

Intern
Intern
User avatar
B
Joined: 15 Sep 2018
Posts: 30
Re: For any positive integer n, the sum of the first n positive integers  [#permalink]

Show Tags

New post 15 Nov 2018, 22:46
We are given the formula:

\(∑_1^n \frac{n(n+1)}{2}\)


We're asked to find sum of all the even numbers between \(99\) and \(301\)....

\(100+102+104+...+296+298+300\)


We can factor out \(2\) from this sum:

\(2 \times (50+51+52+...+148+149+150)\)


In the brackets we have sum of numbers from \(50\) to \(150\). We can use the given formula for the sum of \(1\) to \(150\) and subtract the sum of \(1\) to \(49\) to find the total of this sum.

\(2 \times (\frac{150 \times 151}{2}-\frac{49 \times 50}{2})\)

\(=(150)(151) – (49)(50)\)

\(=22,650 – 2,450\)

\(=20,200\)


The final answer is .
GMAT Club Bot
Re: For any positive integer n, the sum of the first n positive integers &nbs [#permalink] 15 Nov 2018, 22:46

Go to page   Previous    1   2   [ 34 posts ] 

Display posts from previous: Sort by

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

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


Copyright

GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.