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

It is currently 17 Oct 2019, 05:59

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

The sum of the first k positive integers is equal to k(k+1)/2. What is

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

Hide Tags

Find Similar Topics 
Senior Manager
Senior Manager
User avatar
S
Status: No dream is too large, no dreamer is too small
Joined: 14 Jul 2010
Posts: 423
The sum of the first k positive integers is equal to k(k+1)/2. What is  [#permalink]

Show Tags

New post 19 Jan 2012, 11:16
3
25
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

54% (01:55) correct 46% (01:41) wrong based on 361 sessions

HideShow timer Statistics

The sum of the first k positive integers is equal to k(k+1)/2. What is the sum of the integers from n to m, inclusive, where 0<n<m?

A. \(\frac{m(m+1)}{2} - \frac{(n+1)(n+2)}{2}\)

B. \(\frac{m(m+1)}{2} - \frac{n(n+1)}{2}\)

C. \(\frac{m(m+1)}{2} - \frac{(n-1)n}{2}\)

D. \(\frac{(m-1)m}{2} - \frac{(n+1)(n+2)}{2}\)

E. \(\frac{(m-1)m}{2} - \frac{n(n+1)}{2}\)

_________________
Most Helpful Expert Reply
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58417
Re: The sum of the first k positive integers is equal to k(k+1)/2. What is  [#permalink]

Show Tags

New post 19 Jan 2012, 11:28
2
10
Baten80 wrote:
The sum of the first k positive integers is equal to k(k+1)/2. What is the sum of the integers from n to m, inclusive, where 0<n<m?

A. \(\frac{m(m+1)}{2} - \frac{(n+1)(n+2)}{2}\)

B. \(\frac{m(m+1)}{2} - \frac{n(n+1)}{2}\)

C. \(\frac{m(m+1)}{2} - \frac{(n-1)n}{2}\)

D. \(\frac{(m-1)m}{2} - \frac{(n+1)(n+2)}{2}\)

E. \(\frac{(m-1)m}{2} - \frac{n(n+1)}{2}\)


The sum of the integers from n to m, inclusive, will be the sum of the first m positive integers minus the sum of the first n-1 integers: \(\frac{m(m+1)}{2}-\frac{(n-1)(n-1+1)}{2}=\frac{m(m+1)}{2}-\frac{(n-1)n}{2}\).

Answer: C.
_________________
General Discussion
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58417
Re: The sum of the first k positive integers is equal to k(k+1)/2. What is  [#permalink]

Show Tags

New post 19 Jan 2012, 11:34
5
3
Baten80 wrote:
The sum of the first k positive integers is equal to k(k+1)/2. What is the sum of the integers from n to m, inclusive, where 0<n<m?

A. \(\frac{m(m+1)}{2} - \frac{(n+1)(n+2)}{2}\)

B. \(\frac{m(m+1)}{2} - \frac{n(n+1)}{2}\)

C. \(\frac{m(m+1)}{2} - \frac{(n-1)n}{2}\)

D. \(\frac{(m-1)m}{2} - \frac{(n+1)(n+2)}{2}\)

E. \(\frac{(m-1)m}{2} - \frac{n(n+1)}{2}\)



Or try plug-in method: let m=4 and n=3 --> then m+n=7. Let see which option yields 7.

A. \(\frac{m(m+1)}{2} - \frac{(n+1)(n+2)}{2} = 10-10=0\);

B. \(\frac{m(m+1)}{2} - \frac{n(n+1)}{2} = 10-6=4\);

C. \(\frac{m(m+1)}{2} - \frac{(n-1)n}{2} = 10-3=7\) --> OK;

D. \(\frac{(m-1)m}{2} - \frac{(n+1)(n+2)}{2} = 6-10=-4\);

E. \(\frac{(m-1)m}{2} - \frac{n(n+1)}{2} = 6-6=0\).


Answer: C.
_________________
Current Student
avatar
Joined: 03 Apr 2015
Posts: 22
Schools: ISB '16 (A)
Re: The sum of the first k positive integers is equal to k(k+1)/2. What is  [#permalink]

Show Tags

New post 19 Jul 2015, 06:05
Can also be solved by using Sum of A.P formula from n to m. (A bit lengthy though)
Current Student
avatar
B
Joined: 08 Jan 2015
Posts: 77
Location: Thailand
GMAT 1: 540 Q41 V23
GMAT 2: 570 Q44 V24
GMAT 3: 550 Q44 V21
GMAT 4: 660 Q48 V33
GPA: 3.31
WE: Science (Other)
Re: The sum of the first k positive integers is equal to k(k+1)/2. What is  [#permalink]

Show Tags

New post 16 Sep 2015, 21:52
Baten80 wrote:
The sum of the first k positive integers is equal to k(k+1)/2. What is the sum of the integers from n to m, inclusive, where 0<n<m?

A. m(m+1)/2 - (n+1)(n+2)/2
B. m(m+1)/2 - n(n+1)/2
C. m(m+1)/2 - (n-1)n/2
D. (m-1)m/2 - (n+1)(n+2)/2
E. (m-1)m/2 - n(n+1)/2

I got Answer B.
But OA is different.


I think this question can be solved easily by picking numbers.

Let n = 1 and m = 2

Sum of 1 integer is 1;
Sum of 2 integers is 3

So, Sum of the integers from 1 to 2 must be 3. Let's pluck N and M in the choices

A. \(\frac{2(2+1)}{2}\) - \(\frac{(1+1)(1+2)}{2}\) \(= 3 - 3 = 0\)

B. \(\frac{2(2+1)}{2}\) - \(\frac{1(1+1)}{2}\) \(= 3 - 1 = 2\)

C. \(\frac{2(2+1)}{2}\) - \(\frac{(1-1)1}{2}\) \(= 3 - 0 = 3\) Bingo!

D. \(\frac{(2-1)2}{2}\) - \(\frac{(1+1)(1+2)}{2}\) \(= 1 - 3 = -2\)

E. \(\frac{(2-1)2}{2}\) - \(\frac{1(1+1)}{2}\) \(= 1 - 1 = 0\)

Correct me if I'm wrong pls
Board of Directors
User avatar
P
Joined: 17 Jul 2014
Posts: 2510
Location: United States (IL)
Concentration: Finance, Economics
GMAT 1: 650 Q49 V30
GPA: 3.92
WE: General Management (Transportation)
GMAT ToolKit User Reviews Badge
Re: The sum of the first k positive integers is equal to k(k+1)/2. What is  [#permalink]

Show Tags

New post 20 Dec 2015, 18:08
I solved by picking numbers.
n=12
m=15.

only answer choice C yields a valid result.
Intern
Intern
avatar
Joined: 12 Nov 2015
Posts: 43
Re: The sum of the first k positive integers is equal to k(k+1)/2. What is  [#permalink]

Show Tags

New post 20 Dec 2015, 22:14
The only thing to trick here is that we need the sum of (n-1) integers to be subtracted from the sum of m integers.
Manager
Manager
User avatar
S
Joined: 18 Jul 2015
Posts: 71
GMAT 1: 530 Q43 V20
WE: Analyst (Consumer Products)
GMAT ToolKit User
Re: The sum of the first k positive integers is equal to k(k+1)/2. What is  [#permalink]

Show Tags

New post 02 Oct 2018, 02:21
1
Bunuel wrote:
Baten80 wrote:
The sum of the first k positive integers is equal to k(k+1)/2. What is the sum of the integers from n to m, inclusive, where 0<n<m?

A. \(\frac{m(m+1)}{2} - \frac{(n+1)(n+2)}{2}\)

B. \(\frac{m(m+1)}{2} - \frac{n(n+1)}{2}\)

C. \(\frac{m(m+1)}{2} - \frac{(n-1)n}{2}\)

D. \(\frac{(m-1)m}{2} - \frac{(n+1)(n+2)}{2}\)

E. \(\frac{(m-1)m}{2} - \frac{n(n+1)}{2}\)


The sum of the integers from n to m, inclusive, will be the sum of the first m positive integers minus the sum of the first n-1 integers: \(\frac{m(m+1)}{2}-\frac{(n-1)(n-1+1)}{2}=\frac{m(m+1)}{2}-\frac{(n-1)n}{2}\).

Answer: C.


Hi Bunuel. I did not follow the part were you took (n-1). Can you please explain the same. Thanks.
_________________
Cheers. Wishing Luck to Every GMAT Aspirant!
Manager
Manager
User avatar
G
Joined: 09 May 2017
Posts: 232
Location: Iran (Islamic Republic of)
Re: The sum of the first k positive integers is equal to k(k+1)/2. What is  [#permalink]

Show Tags

New post 11 Aug 2019, 04:20
Pritishd wrote:
Bunuel wrote:
Baten80 wrote:
The sum of the first k positive integers is equal to k(k+1)/2. What is the sum of the integers from n to m, inclusive, where 0<n<m?

A. \(\frac{m(m+1)}{2} - \frac{(n+1)(n+2)}{2}\)

B. \(\frac{m(m+1)}{2} - \frac{n(n+1)}{2}\)

C. \(\frac{m(m+1)}{2} - \frac{(n-1)n}{2}\)

D. \(\frac{(m-1)m}{2} - \frac{(n+1)(n+2)}{2}\)

E. \(\frac{(m-1)m}{2} - \frac{n(n+1)}{2}\)


The sum of the integers from n to m, inclusive, will be the sum of the first m positive integers minus the sum of the first n-1 integers: \(\frac{m(m+1)}{2}-\frac{(n-1)(n-1+1)}{2}=\frac{m(m+1)}{2}-\frac{(n-1)n}{2}\).

Answer: C.


Hi Bunuel. I did not follow the part were you took (n-1). Can you please explain the same. Thanks.

good question maybe someday Bunuel answer that :please
_________________
behind every principle is not always promising
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58417
Re: The sum of the first k positive integers is equal to k(k+1)/2. What is  [#permalink]

Show Tags

New post 11 Aug 2019, 04:24
09173140521 wrote:
Pritishd wrote:
Bunuel wrote:
The sum of the first k positive integers is equal to k(k+1)/2. What is the sum of the integers from n to m, inclusive, where 0<n<m?

A. \(\frac{m(m+1)}{2} - \frac{(n+1)(n+2)}{2}\)

B. \(\frac{m(m+1)}{2} - \frac{n(n+1)}{2}\)

C. \(\frac{m(m+1)}{2} - \frac{(n-1)n}{2}\)

D. \(\frac{(m-1)m}{2} - \frac{(n+1)(n+2)}{2}\)

E. \(\frac{(m-1)m}{2} - \frac{n(n+1)}{2}\)

The sum of the integers from n to m, inclusive, will be the sum of the first m positive integers minus the sum of the first n-1 integers: \(\frac{m(m+1)}{2}-\frac{(n-1)(n-1+1)}{2}=\frac{m(m+1)}{2}-\frac{(n-1)n}{2}\).

Answer: C.


Hi Bunuel. I did not follow the part were you took (n-1). Can you please explain the same. Thanks.

good question maybe someday Bunuel answer that :please


_________________________
Not sure what to explain there.
_________________
Manager
Manager
User avatar
G
Joined: 09 May 2017
Posts: 232
Location: Iran (Islamic Republic of)
Re: The sum of the first k positive integers is equal to k(k+1)/2. What is  [#permalink]

Show Tags

New post 11 Aug 2019, 04:34
I GO FOR B but wrong answer .....why we must consider (n-1) although I consider just (n)? cannot find what logic behind that
_________________
behind every principle is not always promising
Manager
Manager
User avatar
S
Joined: 18 Jul 2015
Posts: 71
GMAT 1: 530 Q43 V20
WE: Analyst (Consumer Products)
GMAT ToolKit User
Re: The sum of the first k positive integers is equal to k(k+1)/2. What is  [#permalink]

Show Tags

New post 16 Aug 2019, 04:36
1
09173140521 wrote:
I GO FOR B but wrong answer .....why we must consider (n-1) although I consider just (n)? cannot find what logic behind that


I seem to have an answer to my own question. If you read the question stem closely then it asks for sum from \(n\) to \(m\), inclusive. The word inclusive is the key here. If you take \(n\) then you make the sum exclusive of \(n\), but if you take \((n-1)\) then you make the sum inclusive.

Bunuel - Please correct if my understanding is incorrect. Thanks for the great explanation in your initial post!
_________________
Cheers. Wishing Luck to Every GMAT Aspirant!
Manager
Manager
User avatar
G
Joined: 09 May 2017
Posts: 232
Location: Iran (Islamic Republic of)
Re: The sum of the first k positive integers is equal to k(k+1)/2. What is  [#permalink]

Show Tags

New post 16 Aug 2019, 04:43
Pritishd wrote:
09173140521 wrote:
I GO FOR B but wrong answer .....why we must consider (n-1) although I consider just (n)? cannot find what logic behind that


I seem to have an answer to my own question. If you read the question stem closely then it asks for sum from \(n\) to \(m\), inclusive. The word inclusive is the key here. If you take \(n\) then you make the sum exclusive of \(n\), but if you take \((n-1)\) then you make the sum inclusive.

Bunuel - Please correct if my understanding is incorrect. Thanks for the great explanation in your initial post!
yes it is correct ! ...the key is <inclusive> ....many thanks
_________________
behind every principle is not always promising
SVP
SVP
User avatar
P
Joined: 03 Jun 2019
Posts: 1698
Location: India
Premium Member Reviews Badge CAT Tests
Re: The sum of the first k positive integers is equal to k(k+1)/2. What is  [#permalink]

Show Tags

New post 16 Aug 2019, 04:45
1
Baten80 wrote:
The sum of the first k positive integers is equal to k(k+1)/2. What is the sum of the integers from n to m, inclusive, where 0<n<m?

A. \(\frac{m(m+1)}{2} - \frac{(n+1)(n+2)}{2}\)

B. \(\frac{m(m+1)}{2} - \frac{n(n+1)}{2}\)

C. \(\frac{m(m+1)}{2} - \frac{(n-1)n}{2}\)

D. \(\frac{(m-1)m}{2} - \frac{(n+1)(n+2)}{2}\)

E. \(\frac{(m-1)m}{2} - \frac{n(n+1)}{2}\)


Given: The sum of the first k positive integers is equal to k(k+1)/2.

Asked: What is the sum of the integers from n to m, inclusive, where 0<n<m?

Sum of the integers from n to m, inclusive, where 0<n<m = Sum of first m integers - Sum of first (n-1) integers
= \(\frac{m(m+1)}{2} - \frac{(n-1)n}{2}\)

IMO C
_________________
"Success is not final; failure is not fatal: It is the courage to continue that counts."

Please provide kudos if you like my post. Kudos encourage active discussions.

My GMAT Resources: -

Efficient Learning
All you need to know about GMAT quant

Tele: +91-11-40396815
Mobile : +91-9910661622
E-mail : kinshook.chaturvedi@gmail.com
GMAT Club Bot
Re: The sum of the first k positive integers is equal to k(k+1)/2. What is   [#permalink] 16 Aug 2019, 04:45
Display posts from previous: Sort by

The sum of the first k positive integers is equal to k(k+1)/2. What is

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





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