Author 
Message 
TAGS:

Hide Tags

Retired Moderator
Joined: 29 Oct 2013
Posts: 272
Concentration: Finance
GPA: 3.7
WE: Corporate Finance (Retail Banking)

The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n1)d is giv
[#permalink]
Show Tags
03 Jun 2014, 21:30
Question Stats:
63% (02:31) correct 37% (02:49) wrong based on 459 sessions
HideShow timer Statistics
The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n1)d is give by Sn = n/2(2a + (n1)*d) = . What is the sum of the integers 1 to 100 inclusive, with the even integers between 25 and 63 omitted. A. 4345 B. 4302 C. 4258 D. 4214 E. 4170
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
Please contact me for super inexpensive quality private tutoring
My journey V46 and 750 > http://gmatclub.com/forum/myjourneyto46onverbal750overall171722.html#p1367876




Retired Moderator
Joined: 29 Oct 2013
Posts: 272
Concentration: Finance
GPA: 3.7
WE: Corporate Finance (Retail Banking)

Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n1)d is giv
[#permalink]
Show Tags
03 Jun 2014, 21:36
I think it is faster to ignore the given formula all together. Lets use most commonly known formula for sum of consecutive/evenly spaced numbers: (average of 1st and last no.s)*no of terms Sum of all integers 1 to 100 inclusive : (1+100)/2 * 100 = 5050 Sum of even no.s from 25 to 63: (26+62)/2 * 19 = 836 Required sum : 5050  836 = 4214 Ans: D
_________________
Please contact me for super inexpensive quality private tutoring
My journey V46 and 750 > http://gmatclub.com/forum/myjourneyto46onverbal750overall171722.html#p1367876




SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1834
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)

Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n1)d is giv
[#permalink]
Show Tags
05 Jun 2014, 00:56
We require to find: (1 + 2 + 3 + 4........... + 99 + 100)  (26 + 28 + 30 + ............ + 60 + 62) Expression First(1 + 2 + 3 + 4........... + 99 + 100) \(= \frac{100}{2} * (100 + 1) = 50 * 101 = 5050\) Expression Second(26 + 28 + 30 + ............ + 60 + 62) \(Total terms = \frac{6226}{2} + 1 = 18 + 1 = 19\) \(Addition = \frac{19}{2} (26 + 62) = 44 * 19 = 836\) Answer = 5050  836 = 4214
_________________
Kindly press "+1 Kudos" to appreciate



Intern
Joined: 02 Jul 2014
Posts: 1

Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n1)d is giv
[#permalink]
Show Tags
02 Jul 2014, 15:30
I don't think that on the test, I would know to change 63 and 25 to 62 and 26. If you try to find the number of terms using 63 and 25, you wind up with 20:
[(63  25)/2] + 1 = 20. And 22 times the average of 63 and 25 is 880, which if subtracted from 5,050 would lead you to answer choice D.
How do you recognize this trap and avoid falling into it?



Intern
Joined: 10 Oct 2014
Posts: 1

Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n1)d is giv
[#permalink]
Show Tags
10 Oct 2014, 13:22
for the 2nd part of the problem, omitting the even integers 2563, why does the gmat explanation assign d=2? I understand assigning a=26 and n=19 but not following why d=2. Any assistance would be appreciated.
thanks



Manager
Status: I am not a product of my circumstances. I am a product of my decisions
Joined: 20 Jan 2013
Posts: 124
Location: India
Concentration: Operations, General Management
GPA: 3.92
WE: Operations (Energy and Utilities)

Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n1)d is giv
[#permalink]
Show Tags
11 Oct 2014, 00:15
WishfulThinking1983 wrote: I don't think that on the test, I would know to change 63 and 25 to 62 and 26. If you try to find the number of terms using 63 and 25, you wind up with 20:
[(63  25)/2] + 1 = 20. And 22 times the average of 63 and 25 is 880, which if subtracted from 5,050 would lead you to answer choice D.
How do you recognize this trap and avoid falling into it? By recognizing the fact that we have to subtract the sum of EVEN numbers between 25 and 63 from the sum of all natural numbers from 1 to 100, inclusive.



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8293
Location: Pune, India

Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n1)d is giv
[#permalink]
Show Tags
20 Oct 2015, 21:28
jainaba wrote: for the 2nd part of the problem, omitting the even integers 2563, why does the gmat explanation assign d=2? I understand assigning a=26 and n=19 but not following why d=2. Any assistance would be appreciated.
thanks The first part of the problem is straight forward: Sum of numbers from 1 to 100 = 100*101/2 = 50*101 = 5050 In the second part, you need this sum 26 + 28 + 30 + ... + 62 This is an arithmetic progression with common difference of 2 (between consecutive terms, there is a difference of 2). 26 + (26+2) + (26+2*2) + .... (26+2*18) a, a+d, a+2d,..., a+(n1)d So n1 = 18 and hence, n = 19 Using the formula given, you get Sum = n/2(2a + (n1)*d) = 19/2 * [2*26 + (19 1)*2] = 836 So the required sum = 5050  836 = 4214 For more on arithmetic progressions, check: http://www.veritasprep.com/blog/2012/03 ... gressions/
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!



Senior Manager
Joined: 20 Aug 2015
Posts: 392
Location: India

Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n1)d is giv
[#permalink]
Show Tags
21 Oct 2015, 03:23
MensaNumber wrote: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n1)d is give by Sn = (n/2)*(2a + n1)d. What is the sum of the integers 1 to 100 inclusive, with the even integers between 25 and 63 omitted.
A. 4345 B. 4302 C. 4258 D. 4214 E. 4170 There are two parts to this question: 1st part: Sum of integers from 1 to 100 inclusive We can find this by applying the formula Sn = (n/2)*(2a + n1)d.The other way to write this formula is (n/2)*(a + l), where a = first term and l = last termApplying the formula, Sn = (100/2)(1+100) = 50*101 = 5050 Second Part: Sum of even integers between 25 and 63 The even integers start from 26 and end at 62 No of terms = 19 Hence the sum = (19/2)*(26 + 62) = 836 Subtracting this from 5050, we get 4214. Option D



Manager
Joined: 28 Dec 2013
Posts: 68

Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n1)d is giv
[#permalink]
Show Tags
15 Dec 2015, 15:39
NoHalfMeasures wrote: I think it is faster to ignore the given formula all together.
Lets use most commonly know formula for sum of consecutive/evenly spaced numbers: (average of 1st and last no.s)*no of terms Sum of all integers 1 to 100 inclusive : (1+100)/2 * 100 = 5050 Sum of even no.s from 25 to 63: (26+62)/2 * 19 = 836 Required sum : 5050  836 = 4214 Ans: D where does the 19 come from in sum of evens? and why 62 and not 64?



Intern
Joined: 12 Nov 2015
Posts: 48

Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n1)d is giv
[#permalink]
Show Tags
16 Dec 2015, 02:28
Took 1.5 minutes. D is the correct one.
1100 integer sum= 5050 2563 even integer sum= 836
difference = 4214



Math Expert
Joined: 02 Aug 2009
Posts: 6808

Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n1)d is giv
[#permalink]
Show Tags
16 Dec 2015, 02:47
NoHalfMeasures wrote: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n1)d is give by Sn = (n/2)*(2a + n1)d. What is the sum of the integers 1 to 100 inclusive, with the even integers between 25 and 63 omitted.
A. 4345 B. 4302 C. 4258 D. 4214 E. 4170 Hi, a method to do it under one minutes is to take advantage of the choices given... lets work on the last digit as we have different units digit in each choice...
total sum of 1 to 100 inclusive will have 0 as the last digit.. this is so because the resultant will be 10*(sum of all single digits)... and since we are multiplying by 10, units digit will be 0...
now for single digit in sum of even number from 25 to 63.. 25 to 65 will have 4 times sum of single digit even int, 4*(2+4+6+8+0)=4*20.. here too the units digit is 0, but 64 has to be excluded from the total..
two ways from here on.. 1) we are subtracting 0 from 0 so units digit should be 0, but we have to add 64.. so last/units digit =4..
2)we subtract 64 from sum of even int.. so units digit=804=76.. or units digit =6...
so our answer should have units digit as 106=4.. only D has 4 as units digit.. ans D
_________________
1) Absolute modulus : http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effectsofarithmeticoperationsonfractions269413.html
GMAT online Tutor



Current Student
Joined: 18 Jul 2015
Posts: 41
Location: Brazil
Concentration: General Management, Strategy
GMAT 1: 640 Q39 V38 GMAT 2: 700 Q47 V38

Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n1)d is giv
[#permalink]
Show Tags
19 Feb 2016, 10:44
I got a question!
To find the # of even terms we have to divide by 2.
What if we have a ODD numbers of terms? What do we do?



Manager
Joined: 20 Jan 2016
Posts: 95
Location: Canada
WE: Consulting (Other)

Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n1)d is giv
[#permalink]
Show Tags
28 Sep 2016, 14:19
VeritasPrepKarishma wrote: jainaba wrote: for the 2nd part of the problem, omitting the even integers 2563, why does the gmat explanation assign d=2? I understand assigning a=26 and n=19 but not following why d=2. Any assistance would be appreciated.
thanks The first part of the problem is straight forward: Sum of numbers from 1 to 100 = 100*101/2 = 50*101 = 5050 In the second part, you need this sum 26 + 28 + 30 + ... + 62 This is an arithmetic progression with common difference of 2 (between consecutive terms, there is a difference of 2). 26 + (26+2) + (26+2*2) + .... (26+2*18) a, a+d, a+2d,..., a+(n1)d So n1 = 18 and hence, n = 19 Using the formula given, you get Sum = n/2(2a + (n1)*d) = 19/2 * [2*26 + (19 1)*2] = 836 So the required sum = 5050  836 = 4214 For more on arithmetic progressions, check: http://www.veritasprep.com/blog/2012/03 ... gressions/I think I am missing something here: The formula provided in the question is (n/2)*(2a + n1)d which is different from the formula Karishma provided which is n/2(2a + (n1)*d). if you apply the formula on each you get different results 1) Given Formula (n/2)*(2a + n1)d for even integers between 25 and 63 n=19 d=2 a=26 = (19/2)*(2*26 + 18)2 = (19/2)*(52+18)2 = (19/2)*(70)2 = 19*70 = 1330 2) Karishma's forumla n/2(2a + (n1)*d) for even integers between 25 and 63 n=19 d=2 a=26 = (19/2)(2(26)+(191)2) = 19(26+18) = 19*44 = 836 I am sure my mistake is really obvious but I just can't see it. I would appreciate it if you can correct me where I am wrong.
_________________
Migatte no Gokui



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8293
Location: Pune, India

Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n1)d is giv
[#permalink]
Show Tags
28 Sep 2016, 20:46
oasis90 wrote: VeritasPrepKarishma wrote: jainaba wrote: for the 2nd part of the problem, omitting the even integers 2563, why does the gmat explanation assign d=2? I understand assigning a=26 and n=19 but not following why d=2. Any assistance would be appreciated.
thanks The first part of the problem is straight forward: Sum of numbers from 1 to 100 = 100*101/2 = 50*101 = 5050 In the second part, you need this sum 26 + 28 + 30 + ... + 62 This is an arithmetic progression with common difference of 2 (between consecutive terms, there is a difference of 2). 26 + (26+2) + (26+2*2) + .... (26+2*18) a, a+d, a+2d,..., a+(n1)d So n1 = 18 and hence, n = 19 Using the formula given, you get Sum = n/2(2a + (n1)*d) = 19/2 * [2*26 + (19 1)*2] = 836 So the required sum = 5050  836 = 4214 For more on arithmetic progressions, check: http://www.veritasprep.com/blog/2012/03 ... gressions/I think I am missing something here: The formula provided in the question is (n/2)*(2a + n1)d which is different from the formula Karishma provided which is n/2(2a + (n1)*d). if you apply the formula on each you get different results 1) Given Formula (n/2)*(2a + n1)d for even integers between 25 and 63 n=19 d=2 a=26 = (19/2)*(2*26 + 18)2 = (19/2)*(52+18)2 = (19/2)*(70)2 = 19*70 = 1330 2) Karishma's forumla n/2(2a + (n1)*d) for even integers between 25 and 63 n=19 d=2 a=26 = (19/2)(2(26)+(191)2) = 19(26+18) = 19*44 = 836 I am sure my mistake is really obvious but I just can't see it. I would appreciate it if you can correct me where I am wrong. The OP has made a mistake while writing down the formula in the question. The correct formula is what I have provided. For more on AP formulas, check: http://www.veritasprep.com/blog/2012/03 ... gressions/
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!



Intern
Joined: 11 Jan 2013
Posts: 28

Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n1)d is giv
[#permalink]
Show Tags
06 Feb 2017, 18:05
NoHalfMeasures wrote: I think it is faster to ignore the given formula all together.
Lets use most commonly known formula for sum of consecutive/evenly spaced numbers: (average of 1st and last no.s)*no of terms Sum of all integers 1 to 100 inclusive : (1+100)/2 * 100 = 5050 Sum of even no.s from 25 to 63: (26+62)/2 * 19 = 836 Required sum : 5050  836 = 4214 Ans: D I stared at this question and got super confused by the formula. Thank you for clarifying that you can just ignore that. I could have solved this one!



Senior Manager
Status: Come! Fall in Love with Learning!
Joined: 05 Jan 2017
Posts: 457
Location: India

Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n1)d is giv
[#permalink]
Show Tags
01 Mar 2017, 01:08
Sum = sum from 1 to 100  sum from 26 to 62 = 100(100+1)/2  19(26 +62)/2 = 5050  836 = 4214
_________________
GMAT Mentors



Intern
Joined: 04 Feb 2017
Posts: 1

Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n1)d is giv
[#permalink]
Show Tags
18 Jul 2017, 06:34
Could someone tell me how did you know that (2a+(n1)d) was referring to first plus last one? And what is d?
Thanks



Target Test Prep Representative
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2835

Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n1)d is giv
[#permalink]
Show Tags
20 Jul 2017, 16:45
NoHalfMeasures wrote: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n1)d is give by Sn = n/2(2a + (n1)*d) = . What is the sum of the integers 1 to 100 inclusive, with the even integers between 25 and 63 omitted.
A. 4345 B. 4302 C. 4258 D. 4214 E. 4170 We can find the sum of the integers 1 to 100 inclusive and subtract from it the sum of the even integers between 25 and 63 (i.e., the even integers from 26 to 62 inclusive). Using the formula given in the question for the sum of the integers 1 to 100 inclusive, we let a = 1, d = 1, and n = 100; thus, the sum is: 100/2[2(1) + (100  1)(1)] = 50(101) = 5050 To calculate the sum of the even integers 26 to 62 inclusive, we let a = 26, d = 2, and n = (62  26)/2 + 1 = 19; thus, the sum is: 19/2[26(2) + (19  1)(2)] = 19/2(88) = 19(44) = 836 Therefore, the sum of the integers from 1 to 100 inclusive, with the sum of the even integers between 25 and 63 omitted, is: 5050  836 = 4214 Note: We don’t need to use the formula given in the problem; we can use the basic formula for the sum of an evenly spaced set of numbers, sum = quantity x average. For example, for the sum of the integers 1 to 100 inclusive, quantity = 100 and average = (1 + 100)/2 = 50.5. Thus, sum = 100 x 50.5 = 5050. Answer: D
_________________
Jeffery Miller
Head of GMAT Instruction
GMAT Quant SelfStudy Course
500+ lessons 3000+ practice problems 800+ HD solutions



Intern
Joined: 01 Aug 2018
Posts: 3

Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n1)d is giv
[#permalink]
Show Tags
27 Aug 2018, 12:02
Hi, I do not understand why we substract 836 from 5050. Could anyone help me?
Thank you, Amanda




Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n1)d is giv &nbs
[#permalink]
27 Aug 2018, 12:02






