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The sum of the first 50 positive even integers is 2550. What [#permalink]
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03 Sep 2009, 06:35
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The sum of the first 50 positive even integers is 2550. What is the sum of even integers from 102 to 200 inclusive? A. 5100 B. 7550 C. 10100 D. 15500 E. 20100 OPEN DISCUSSION OF THIS QUESTION IS HERE: thesumofthefirst50positiveevenintegersis2550what129468.html
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Last edited by Bunuel on 26 Mar 2014, 01:42, edited 3 times in total.
Renamed the topic and edited the question.



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Re: GMATPrep: sum of even integers from 102 to 200 [#permalink]
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03 Sep 2009, 06:43
Can you please explain how this is solved
Thanks



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Re: GMATPrep: sum of even integers from 102 to 200 [#permalink]
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03 Sep 2009, 06:44
Can someone tell me how do we need to approach that kind of question?
Rgds



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Re: GMATPrep: sum of even integers from 102 to 200 [#permalink]
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03 Sep 2009, 06:44
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My solution is:
First 50 even integers: 2 4 6 8 <...>
Integers from 102 to 200 102 104 106 108 <...>
We notice that each integer from the second set is 100 more than the respective integer in the first set. Since we have 50 even integers from 102 to 200, then: 2550+(100*50)=7550.
Are there any other ways to solve the prob? Thanks!
_________________________________ Please kudos me if you find my post useful



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Re: GMATPrep: sum of even integers from 102 to 200 [#permalink]
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03 Sep 2009, 06:53
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CasperMonday wrote: The sum of the first 50 positive even integers is 2550. What is the sum of even integers from 102 to 200 inclusive? a. 5100 b. 7550 c. 10100 d. 15500 e. 20100 OA I had derived a formula for calculating the summation of even/odd numbers in a series, when I was in high school. Sum of EVEN numbers in a series from 1 > n (where n is the LAST EVEN number in a series) = n*(n+2)/4 Sum of ODD numbers in a series from 1 > n (where n is the LAST ODD number in a series) = (n+1)*(n+1)/4 This works perfect for any series starting from 1. Using the above formula we can solve the problem: 200*202/4  100*102/4 = 7550 Please give Kudos if this was helpful.



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Re: GMATPrep: sum of even integers from 102 to 200 [#permalink]
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03 Sep 2009, 06:56
It sure is. But can we 'trust' this formula? Can it be so that it is only applicable to certain cases and there are exceptions? (besides what you mentioned?)



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Re: GMATPrep: sum of even integers from 102 to 200 [#permalink]
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03 Sep 2009, 06:58
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CasperMonday wrote: The sum of the first 50 positive even integers is 2550. What is the sum of even integers from 102 to 200 inclusive? a. 5100 b. 7550 c. 10100 d. 15500 e. 20100 OA Sum of even integers from 102200(both included)No of terms = (200102)/2 + 1 = 50 (+1 because both are included) If you observe, this is AP (Arithmetic progression): 102,104,106,......,200 Sum of n terms in AP Formula, Sum = n/2[2a + (n1)d] where, n = no of terms = 50 in this case. a = starting number = 102 d = difference between AP = 2 So, Sum = 50/2[2*102 + (501)2] Solving this, we can get the answer. This is one more solution for this kind of problems.



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Re: GMATPrep: sum of even integers from 102 to 200 [#permalink]
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03 Sep 2009, 07:02
CasperMonday wrote: It sure is. But can we 'trust' this formula? Can it be so that it is only applicable to certain cases and there are exceptions? (besides what you mentioned?) As far as I know and I have seen in similar problems, using Arithmetic Progression (AP) formulas is generally used. For eg: sum of even nos from 1001000 sum of odd nos from 3999 Sum of nos divisible by 5 from 1005000 In all the cases above, common difference and starting and end point is given. So, AP is a sure shot way to approach this and trustable also.



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Re: GMATPrep: sum of even integers from 102 to 200 [#permalink]
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03 Sep 2009, 07:10
Greatly appreciated)



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Re: GMATPrep: sum of even integers from 102 to 200 [#permalink]
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03 Sep 2009, 07:10
CasperMonday wrote: It sure is. But can we 'trust' this formula? Can it be so that it is only applicable to certain cases and there are exceptions? (besides what you mentioned?) At least I have not yet found any exception in last 12 years. As far as the series starts from 1, and we consider the last ODD/EVEN number of the series correctly, it works.



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Re: GMATPrep: sum of even integers from 102 to 200 [#permalink]
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03 Sep 2009, 07:49
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You know what came into my mind? How would a CR question be formulated))) Like, which question will be best to ask to assess reasonability of the statement? One of the possible answers: "how many arithm progression problems did you solve during the last 12 years"?)
Well, I am spamming now and better stop)



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Re: GMATPrep: sum of even integers from 102 to 200 [#permalink]
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03 Sep 2009, 08:23
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Sum of even integers from 102200(both included)
No of terms = (200102)/2 + 1 = 50 (+1 because both are included)
If you observe, this is AP (Arithmetic progression): 102,104,106,......,200 Sum of n terms in AP Formula, Sum = n/2[2a + (n1)d] where, n = no of terms = 50 in this case. a = starting number = 102 d = difference between AP = 2
So, Sum = 50/2[2*102 + (501)2]
Solving this, we can get the answer.
This is one more solution for this kind of problems.
Saurabhricha, what would be the formula for odd nbrs?



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Re: GMATPrep: sum of even integers from 102 to 200 [#permalink]
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03 Sep 2009, 10:01
CasperMonday wrote: You know what came into my mind? How would a CR question be formulated))) Like, which question will be best to ask to assess reasonability of the statement? One of the possible answers: "how many arithm progression problems did you solve during the last 12 years"?)
Well, I am spamming now and better stop) +1



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Re: GMATPrep: sum of even integers from 102 to 200 [#permalink]
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03 Sep 2009, 10:54
hey,
can someone please solve the eq for me... i guess i am misunderstanding some of the symbols and my answer is not coinciding with the correct one...



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Re: GMATPrep: sum of even integers from 102 to 200 [#permalink]
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03 Sep 2009, 11:15
defoue wrote: Saurabhricha, what would be the formula for odd nbrs? defoue, The solution and formula is not only for even numbers. It holds good for odd numbers as well. The only difference in the equation will be the starting point "a" which will be an odd number.



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Re: GMATPrep: sum of even integers from 102 to 200 [#permalink]
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03 Sep 2009, 11:18
meenal8284 wrote: hey,
can someone please solve the eq for me... i guess i am misunderstanding some of the symbols and my answer is not coinciding with the correct one... meenal8284, n = no of terms = 50 in this case. a = starting number = 102 d = difference between AP = 2 So, Sum = 50/2[2*102 + (501)2] sum = 25[204 + 49*2] sum = 25[204 + 98] sum = 25*302 = 7550 I hope it clarifies your doubt regarding calculations related to this problem.



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Re: GMATPrep: sum of even integers from 102 to 200 [#permalink]
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03 Sep 2009, 21:23
saurabhricha wrote: CasperMonday wrote: It sure is. But can we 'trust' this formula? Can it be so that it is only applicable to certain cases and there are exceptions? (besides what you mentioned?) As far as I know and I have seen in similar problems, using Arithmetic Progression (AP) formulas is generally used. For eg: sum of even nos from 1001000 sum of odd nos from 3999 Sum of nos divisible by 5 from 1005000 In all the cases above, common difference and starting and end point is given. So, AP is a sure shot way to approach this and trustable also. There is a generalized formula for the sum of even and odd numbers: For Odd numbers :1,3,5,7.....(2n1) = n^2 For even numbers: 2,4,6,.......2n = n(n+1) So with that in mind the question becomes a lot more simple. Frm : 2, 4....200 ..the sum would be : 100 *101A Frm: 2, 4....100 (already given in question) Sum would be = 50*51B Ans = AB



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Re: GMATPrep: sum of even integers from 102 to 200 [#permalink]
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15 Sep 2009, 10:51
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Here is another way of solving the problem.
Rule: the sum of even numbers between 2,..., n may be found by applying this shortcut: (n/2)*(n/2 + 1).
So the sum of 2,...,200 = 100*101 = 10,100. Taking away the sum of 2,...,100 = 2,550 You get 7,550.



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Re: GMATPrep: sum of even integers from 102 to 200 [#permalink]
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15 Sep 2009, 11:14
This is an A.P with a = 102 , l = 200 and d = 2, n = 50
so sum = n/2 ( a+l)
Sum  50/2 ( 302) = 7550
option B



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Re: GMATPrep: sum of even integers from 102 to 200 [#permalink]
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16 Sep 2009, 02:04
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A)To find the number of terms within a set of consecutive integers, inclusive, where the difference between each term is bigger than 1:
[(Last term  first term)/difference between each integer]+ 1
B)This set of consecutive integers is also an arithmetic sequence. To find the sum of consecutive integers in a sequence:
(mean)(number of terms in sequence)
C)To find the mean of a sequence of consecutive integers: (First term + Last term)/2
Hence: 1) Using equation A above: [(200102)/2]+1 = 50. There are 50 even numbers from 102 to 200, inclusive.
2) Using equation C above to find the mean of the arithmetic sequence: (200+102)/2 = 151
3) Using equation b above to find the sum of even integers from 102 to 200 inclusive: (151)(50) = 7550
Answer is B




Re: GMATPrep: sum of even integers from 102 to 200
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