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

 It is currently 16 Dec 2018, 10:00

### 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 December
PrevNext
SuMoTuWeThFrSa
2526272829301
2345678
9101112131415
16171819202122
23242526272829
303112345
Open Detailed Calendar
• ### Free GMAT Prep Hour

December 16, 2018

December 16, 2018

03:00 PM EST

04:00 PM EST

Strategies and techniques for approaching featured GMAT topics
• ### FREE Quant Workshop by e-GMAT!

December 16, 2018

December 16, 2018

07:00 AM PST

09:00 AM PST

Get personalized insights on how to achieve your Target Quant Score.

# Sum of n positive integers formula for consecutive integers?

Author Message
TAGS:

### Hide Tags

Manager
Joined: 11 Sep 2012
Posts: 87
Sum of n positive integers formula for consecutive integers?  [#permalink]

### Show Tags

16 Nov 2013, 16:15
7
Is the sum of n positive integers formula, $$\frac{n(n+1)}{2}$$, applicable to sets of consecutive integers as well?

For instance, if I wanted to calculate sum of first 20 multiples of 3, could I use: 3 x $$\frac{20(21)}{2}$$

In OG PS Q172, a similar approach has been used for even numbers.
VP
Joined: 02 Jul 2012
Posts: 1176
Location: India
Concentration: Strategy
GMAT 1: 740 Q49 V42
GPA: 3.8
WE: Engineering (Energy and Utilities)
Re: Sum of n positive integers formula for consecutive integers?  [#permalink]

### Show Tags

17 Nov 2013, 00:27
1
1
3+6+9+12+15+18+21+24+27+30...... = 3*(1+2+3+4+5+6+7+8+9+10...)

So, yup.. You can do that...
_________________

Did you find this post helpful?... Please let me know through the Kudos button.

Thanks To The Almighty - My GMAT Debrief

GMAT Reading Comprehension: 7 Most Common Passage Types

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8678
Location: Pune, India
Re: Sum of n positive integers formula for consecutive integers?  [#permalink]

### Show Tags

17 Nov 2013, 19:25
11
10
bschoolaspirant9 wrote:
Is the sum of n positive integers formula, $$\frac{n(n+1)}{2}$$, applicable to sets of consecutive integers as well?

For instance, if I wanted to calculate sum of first 20 multiples of 3, could I use: 3 x $$\frac{20(21)}{2}$$

In OG PS Q172, a similar approach has been used for even numbers.

Note that the formula n(n+1)/2 is used only for adding n consecutive integers starting from 1.
A problem may not directly ask you for this but if you can break it down such that you have to find the sum of 'n consecutive integers starting from 1' then you can use this formula.

In the even integers questions, you may be required to find the sum of first 10 even integers.
2 + 4 + 6 + ... + 18 + 20
Take 2 common, 2*(1 + 2 + 3 + ...10)
To find the sum of the highlighted part, we can use the formula. Then we can multiply it by 2 to get the required sum.

Note that for odd integers, you cannot directly use this formula.
Sum the first 10 odd integers
1 + 3 + 5 + 7+...+19
But you can still make some modifications to find the sum.

1 + 3 + 5 + 7+...+19 = (1 +2+ 3 + 4+5 + 6+ 7+...+19 + 20) - (2 + 4+ 6+...20)
We know how to sum consecutive integers.
(1 +2+ 3 + 4+5 + 6+ 7+...+19 + 20) = 20*21/2
(2 + 4+ 6+...20) = 2 * (10*11)/2 = 10*11 (as before)

So 1 + 3 + 5 + 7+...+19 = (20*21/2) - (10*11) = 100

The direct formula of sum of n consecutive odd integers starting from 1 = n^2
_________________

Karishma
Veritas Prep GMAT Instructor

Manager
Joined: 11 Sep 2012
Posts: 87
Re: Sum of n positive integers formula for consecutive integers?  [#permalink]

### Show Tags

17 Nov 2013, 21:29
1
VeritasPrepKarishma wrote:
bschoolaspirant9 wrote:
Is the sum of n positive integers formula, $$\frac{n(n+1)}{2}$$, applicable to sets of consecutive integers as well?

For instance, if I wanted to calculate sum of first 20 multiples of 3, could I use: 3 x $$\frac{20(21)}{2}$$

In OG PS Q172, a similar approach has been used for even numbers.

Note that the formula n(n+1)/2 is used only for adding n consecutive integers starting from 1.
A problem may not directly ask you for this but if you can break it down such that you have to find the sum of 'n consecutive integers starting from 1' then you can use this formula.

In the even integers questions, you may be required to find the sum of first 10 even integers.
2 + 4 + 6 + ... + 18 + 20
Take 2 common, 2*(1 + 2 + 3 + ...10)
To find the sum of the highlighted part, we can use the formula. Then we can multiply it by 2 to get the required sum.

Note that for odd integers, you cannot directly use this formula.
Sum the first 10 odd integers
1 + 3 + 5 + 7+...+19
But you can still make some modifications to find the sum.

1 + 3 + 5 + 7+...+19 = (1 +2+ 3 + 4+5 + 6+ 7+...+19 + 20) - (2 + 4+ 6+...20)
We know how to sum consecutive integers.
(1 +2+ 3 + 4+5 + 6+ 7+...+19 + 20) = 20*21/2
(2 + 4+ 6+...20) = 2 * (10*11)/2 = 10*11 (as before)

So 1 + 3 + 5 + 7+...+19 = (20*21/2) - (10*11) = 100

The direct formula of sum of n consecutive odd integers starting from 1 = n^2

Thank you for the detailed response.
Intern
Joined: 21 Dec 2014
Posts: 6
Re: Sum of n positive integers formula for consecutive integers?  [#permalink]

### Show Tags

11 Jan 2015, 10:17
Great help guys, an Karishma, huge contribution, thanks from Chile!
Intern
Joined: 16 Jul 2011
Posts: 40
Concentration: Marketing, Real Estate
GMAT 1: 550 Q37 V28
GMAT 2: 610 Q43 V31
Re: Sum of n positive integers formula for consecutive integers?  [#permalink]

### Show Tags

19 Jun 2015, 07:13
VeritasPrepKarishma wrote:
bschoolaspirant9 wrote:
Is the sum of n positive integers formula, $$\frac{n(n+1)}{2}$$, applicable to sets of consecutive integers as well?

For instance, if I wanted to calculate sum of first 20 multiples of 3, could I use: 3 x $$\frac{20(21)}{2}$$

In OG PS Q172, a similar approach has been used for even numbers.

Note that the formula n(n+1)/2 is used only for adding n consecutive integers starting from 1.
A problem may not directly ask you for this but if you can break it down such that you have to find the sum of 'n consecutive integers starting from 1' then you can use this formula.

In the even integers questions, you may be required to find the sum of first 10 even integers.
2 + 4 + 6 + ... + 18 + 20
Take 2 common, 2*(1 + 2 + 3 + ...10)
To find the sum of the highlighted part, we can use the formula. Then we can multiply it by 2 to get the required sum.

Note that for odd integers, you cannot directly use this formula.
Sum the first 10 odd integers
1 + 3 + 5 + 7+...+19
But you can still make some modifications to find the sum.

1 + 3 + 5 + 7+...+19 = (1 +2+ 3 + 4+5 + 6+ 7+...+19 + 20) - (2 + 4+ 6+...20)
We know how to sum consecutive integers.
(1 +2+ 3 + 4+5 + 6+ 7+...+19 + 20) = 20*21/2
(2 + 4+ 6+...20) = 2 * (10*11)/2 = 10*11 (as before)

So 1 + 3 + 5 + 7+...+19 = (20*21/2) - (10*11) = 100

The direct formula of sum of n consecutive odd integers starting from 1 = n^2

Great explanation. However, I have a suggestion. Why can't we use one single formula for all consecutive evenly spaced numbers which is Sum = (average)(number of terms). I hope i am correct?
_________________

"The fool didn't know it was impossible, so he did it."

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8678
Location: Pune, India
Re: Sum of n positive integers formula for consecutive integers?  [#permalink]

### Show Tags

21 Jun 2015, 20:57
2
samdighe wrote:
VeritasPrepKarishma wrote:
bschoolaspirant9 wrote:
Is the sum of n positive integers formula, $$\frac{n(n+1)}{2}$$, applicable to sets of consecutive integers as well?

For instance, if I wanted to calculate sum of first 20 multiples of 3, could I use: 3 x $$\frac{20(21)}{2}$$

In OG PS Q172, a similar approach has been used for even numbers.

Note that the formula n(n+1)/2 is used only for adding n consecutive integers starting from 1.
A problem may not directly ask you for this but if you can break it down such that you have to find the sum of 'n consecutive integers starting from 1' then you can use this formula.

In the even integers questions, you may be required to find the sum of first 10 even integers.
2 + 4 + 6 + ... + 18 + 20
Take 2 common, 2*(1 + 2 + 3 + ...10)
To find the sum of the highlighted part, we can use the formula. Then we can multiply it by 2 to get the required sum.

Note that for odd integers, you cannot directly use this formula.
Sum the first 10 odd integers
1 + 3 + 5 + 7+...+19
But you can still make some modifications to find the sum.

1 + 3 + 5 + 7+...+19 = (1 +2+ 3 + 4+5 + 6+ 7+...+19 + 20) - (2 + 4+ 6+...20)
We know how to sum consecutive integers.
(1 +2+ 3 + 4+5 + 6+ 7+...+19 + 20) = 20*21/2
(2 + 4+ 6+...20) = 2 * (10*11)/2 = 10*11 (as before)

So 1 + 3 + 5 + 7+...+19 = (20*21/2) - (10*11) = 100

The direct formula of sum of n consecutive odd integers starting from 1 = n^2

Great explanation. However, I have a suggestion. Why can't we use one single formula for all consecutive evenly spaced numbers which is Sum = (average)(number of terms). I hope i am correct?

We can, provided we know the average and the number of terms.
If we are asked to find the sum of first 50 consecutive positive odd integers, it might be easier to use 50^2 than to find average and then find the sum.

Mind you, I myself believe in knowing just the main all-applicable kind of formulas and then twisting them around to apply to any situation. But some people prefer to work more on specific formulas and these discussions are for their benefit.
_________________

Karishma
Veritas Prep GMAT Instructor

Director
Affiliations: GMATQuantum
Joined: 19 Apr 2009
Posts: 609
Re: Sum of n positive integers formula for consecutive integers?  [#permalink]

### Show Tags

24 Jun 2015, 03:57
8
2
The best way to deal with sum of any type of arithmetic series is to use the following expression:

Sum = Average*Number of terms

Average = (First term + Last term)/2

Note: The average or the median can be related to the first and last terms, which are often given for these types of problems.

Number of Terms = (Last Term - First Term)/Spacing + 1

Also, I recommend that you understand the basis of these relationships as opposed to blindly memorizing them.

The above relationships are flexible enough to allow you to deal with any GMAT problem on these concepts.

For example in Q172 from the Official Guide: What is the sum of all the even integers between 99 and 301 ?

Average = (100+300)/2 = 200

Number of Terms = (300-100)/2 + 1 = 101 (Note spacing is 2 between consecutive even integers)

Sum = 200*101=20200

Cheers,
Dabral
Non-Human User
Joined: 09 Sep 2013
Posts: 9185
Re: Sum of n positive integers formula for consecutive integers?  [#permalink]

### Show Tags

04 Oct 2018, 01:32
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.
_________________
Re: Sum of n positive integers formula for consecutive integers? &nbs [#permalink] 04 Oct 2018, 01:32
Display posts from previous: Sort by