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The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n1)d is giv
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03 Jun 2014, 20:30
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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
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Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n1)d is giv
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03 Jun 2014, 20:36
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
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Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n1)d is giv
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04 Jun 2014, 23:56
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
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Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n1)d is giv
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02 Jul 2014, 14:30
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?



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Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n1)d is giv
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10 Oct 2014, 12:22
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



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Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n1)d is giv
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10 Oct 2014, 23:15
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.



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Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n1)d is giv
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20 Oct 2015, 20:28
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/
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Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n1)d is giv
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21 Oct 2015, 02:23
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



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Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n1)d is giv
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15 Dec 2015, 14:39
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?



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Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n1)d is giv
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16 Dec 2015, 01:28
Took 1.5 minutes. D is the correct one.
1100 integer sum= 5050 2563 even integer sum= 836
difference = 4214



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Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n1)d is giv
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16 Dec 2015, 01:47
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
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Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n1)d is giv
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19 Feb 2016, 09:44
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?



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Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n1)d is giv
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28 Sep 2016, 13:19
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.
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Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n1)d is giv
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28 Sep 2016, 19:46
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/
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Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n1)d is giv
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06 Feb 2017, 17:05
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!



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Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n1)d is giv
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01 Mar 2017, 00:08
Sum = sum from 1 to 100  sum from 26 to 62 = 100(100+1)/2  19(26 +62)/2 = 5050  836 = 4214
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Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n1)d is giv
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18 Jul 2017, 05:34
Could someone tell me how did you know that (2a+(n1)d) was referring to first plus last one? And what is d?
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Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n1)d is giv
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20 Jul 2017, 15:45
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
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Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n1)d is giv
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27 Aug 2018, 11:02
Hi, I do not understand why we substract 836 from 5050. Could anyone help me?
Thank you, Amanda




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