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The sum of first \(N\) consecutive positive odd integers is \(N^2\) . What is the sum of all odd integers between 13 and 39, inclusive? A. 351 B. 364 C. 410 D. 424 E. 450
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16 Sep 2014, 01:09
Official Solution:The sum of first \(N\) consecutive positive odd integers is \(N^2\) . What is the sum of all odd integers between 13 and 39, inclusive?A. 351 B. 364 C. 410 D. 424 E. 450 The sum of all odd integers between 13 and 39, inclusive equals to the sum of all odd integers from 1 to 39, inclusive minus the sum of all odd integers from 1 to 11, inclusive. Since there are 20 odd integers from 1 to 39, inclusive then the sum of all odd integers from 1 to 39, inclusive is \(20^2\); Since there are 6 odd integers from 1 to 11, inclusive then the sum of all odd integers from 1 to 11, inclusive is \(6^2\); So, the required sum is \(20^26^2=364\). Answer: B
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Re: M2031
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03 Jan 2016, 04:31
HI. we can restate the problem as : what is the summation of evenly spaced series (arithmatic progression) 13,15,17...,39 no. of terms of the series= (3913)/2 +1=14 and avg of the series =(1st term+last term)/2=52/2=26. so summation =26*14=364



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29 Jun 2017, 12:19
I think this is a highquality question and I don't agree with the explanation. it says between 13 and 39 inclusive, not between 11 and 39.



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29 Jun 2017, 14:46
rer05 wrote: I think this is a highquality question and I don't agree with the explanation. it says between 13 and 39 inclusive, not between 11 and 39. Please reread the solution. It says that the sum of all odd integers between 13 and 39, inclusive equals to the sum of all odd integers from 1 to 39, inclusive minus the sum of all odd integers from 1 to 11, inclusive.
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07 Aug 2017, 03:06
Why are we taking 'the sum of all odd integers from 1 to 39, inclusive ' even though the question asks for odd numbers 'BETWEEN' 13 and 39?



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07 Aug 2017, 16:38
chaitalip wrote: Why are we taking 'the sum of all odd integers from 1 to 39, inclusive ' even though the question asks for odd numbers 'BETWEEN' 13 and 39? Please reread the solution. It says that the sum of all odd integers between 13 and 39, inclusive equals to the sum of all odd integers from 1 to 39, inclusive minus the sum of all odd integers from 1 to 11, inclusive.
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Re: M2031
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07 Aug 2017, 21:53
right my bad. thank you!



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17 Sep 2017, 05:10
Hi Bunuel, Please help me understand this question. As per my reasoning goes, there are total (3913)/2+ 1 odd terms. i.e. N= 14. thus according to the question sum of n is n2 which is 14*14. What is the link am missing here ?



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17 Sep 2017, 05:18
jyotipes21@gmail.com wrote: Hi Bunuel, Please help me understand this question. As per my reasoning goes, there are total (3913)/2+ 1 odd terms. i.e. N= 14. thus according to the question sum of n is n2 which is 14*14. What is the link am missing here ? The question says that: The sum of first N consecutive consecutive odd integers is N^2, not that the sum of any N consecutive integers is N^2.
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Re: M2031
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17 Sep 2017, 09:05
How do we know the sum of all the integers 1 to 39 is 400? Is that derived a formula, trick or just something that a person should know. What about other numbers or evens? Thanks.
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17 Sep 2017, 09:27
Plunkster82 wrote: How do we know the sum of all the integers 1 to 39 is 400? Is that derived a formula, trick or just something that a person should know. What about other numbers or evens?
Thanks. The sum of 20 odd integers from 1 to 39 (20 integers) is 20^2 = 400. You can derive it from arithmetic progression formula: https://gmatclub.com/forum/mathsequenc ... 01891.html
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Re: M2031
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26 Oct 2017, 07:52
Great explanations above. This is how I solved it .. Sum of all odd numbers from 13 to 39 is ((14/2) * (13+39)) This comes to 364. In other words, we are concerned about the sum of all numbers in an AP with a CD of 2 and the first term as 13. The answer is option B
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13 Dec 2017, 01:22
I used the formulae :n/2 * 2a+(n1)d



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\(\frac{3913}{2}=13+1=14\) \(\frac{14}{2} = 7\) numbers so then: 13 / 15 / 17 / 19 / 21 / 23 / 25 / Average / ...other 7 odd consecutive numbers Average= 26 Sum: 14*26 = 364 > Ans B
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09 Feb 2019, 11:42
Hi BunuelI don't get how you arrive to the conclusion that "It says that the sum of all odd integers between 13 and 39, inclusive equals to the sum of all odd integers from 1 to 39, inclusive minus the sum of all odd integers from 1 to 11, inclusive" from the question... would you pls further explain?
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10 Feb 2019, 01:46
patto wrote: Hi BunuelI don't get how you arrive to the conclusion that "It says that the sum of all odd integers between 13 and 39, inclusive equals to the sum of all odd integers from 1 to 39, inclusive minus the sum of all odd integers from 1 to 11, inclusive" from the question... would you pls further explain? Take a simple example: the sum of all odd integers between 5 and 9, inclusive ( 5 + 7 + 9) equals to the sum of all odd integers from 1 to 9, inclusive ( 1 + 3 + 5 + 7 + 9) minus the sum of all odd integers from 1 to 3, inclusive ( 1 + 3).
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Bunuel wrote: Official Solution:
The sum of first \(N\) consecutive positive odd integers is \(N^2\) . What is the sum of all odd integers between 13 and 39, inclusive?
A. 351 B. 364 C. 410 D. 424 E. 450
The sum of all odd integers between 13 and 39, inclusive equals to the sum of all odd integers from 1 to 39, inclusive[/b] minus the sum of all odd integers from 1 to 11, inclusive. Since there are 20 odd integers from 1 to 39, inclusive then the sum of all odd integers from 1 to 39, inclusive is \(20^2\); Since there are 6 odd integers from 1 to 11, inclusive then the sum of all odd integers from 1 to 11, inclusive is \(6^2\); So, the required sum is \(20^26^2=364\).
Answer: B Hi Bunuel, I am not able to understand this The sum of all odd i ntegers between 13 and 39, inclusive equals to the sum of all odd integers from 1 to 39, inclusive minus [b]the sum of all odd integers from 1 to 11, inclusiveWhere does this come from.Is there any basis, 11 is specific for this case or for every case
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24 Sep 2019, 07:26
Aadi01 wrote: Bunuel wrote: Official Solution:
The sum of first \(N\) consecutive positive odd integers is \(N^2\) . What is the sum of all odd integers between 13 and 39, inclusive?
A. 351 B. 364 C. 410 D. 424 E. 450
The sum of all odd integers between 13 and 39, inclusive equals to the sum of all odd integers from 1 to 39, inclusive[/b] minus the sum of all odd integers from 1 to 11, inclusive. Since there are 20 odd integers from 1 to 39, inclusive then the sum of all odd integers from 1 to 39, inclusive is \(20^2\); Since there are 6 odd integers from 1 to 11, inclusive then the sum of all odd integers from 1 to 11, inclusive is \(6^2\); So, the required sum is \(20^26^2=364\).
Answer: B Hi Bunuel, I am not able to understand this The sum of all odd i ntegers between 13 and 39, inclusive equals to the sum of all odd integers from 1 to 39, inclusive minus [b]the sum of all odd integers from 1 to 11, inclusiveWhere does this come from.Is there any basis, 11 is specific for this case or for every case It's simple logic: 1 + 3 + 5 +7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29 + 31 + 33 + 35 + 37 + 39RED (the sum of all odd integers between 13 and 39, inclusive) equals to TOTAL (the sum of all odd integers from 1 to 39, inclusive) minus BLUE (the sum of all odd integers from 1 to 11, inclusive). Does this make sense?
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