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Math Expert V
Joined: 02 Sep 2009
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Difficulty:   45% (medium)

Question Stats: 72% (02:04) correct 28% (02:18) wrong based on 149 sessions

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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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Math Expert V
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Posts: 58320

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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 $$20^2-6^2=364$$.

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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= (39-13)/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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I think this is a high-quality question and I don't agree with the explanation. it says between 13 and 39 inclusive, not between 11 and 39.
Math Expert V
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rer05 wrote:
I think this is a high-quality question and I don't agree with the explanation. it says between 13 and 39 inclusive, not between 11 and 39.

Please re-read 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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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?
Math Expert V
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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 re-read 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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right my bad. thank you!
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Hi Bunuel,
Please help me understand this question. As per my reasoning goes, there are total (39-13)/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 ?
Math Expert V
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jyotipes21@gmail.com wrote:
Hi Bunuel,
Please help me understand this question. As per my reasoning goes, there are total (39-13)/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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Concentration: Technology
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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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Math Expert V
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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/math-sequenc ... 01891.html
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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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I used the formulae :n/2 * 2a+(n-1)d
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$$\frac{39-13}{2}=13+1=14$$

$$\frac{14}{2} = 7$$ 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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Hi Bunuel

I 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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Math Expert V
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patto wrote:
Hi Bunuel

I 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 $$20^2-6^2=364$$.

Hi Bunuel,
I am not able to understand this 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 [b]the sum of all odd integers from 1 to 11, inclusive
Where does this come from.Is there any basis, 11 is specific for this case or for every case
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Math Expert V
Joined: 02 Sep 2009
Posts: 58320

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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 $$20^2-6^2=364$$.

Hi Bunuel,
I am not able to understand this 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 [b]the sum of all odd integers from 1 to 11, inclusive
Where 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 + 39

RED (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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