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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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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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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
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 \(20^2-6^2=364\).


Answer: B

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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Aadi01
Bunuel
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 \(20^2-6^2=364\).


Answer: B

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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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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[Sum of all odd integers between 13 and 39, inclusive] = [Sum of positive odd integers until 39] - [Sum of positive odd integers until 11]


1 = 0 * 2 + 1 (=> 1st positive odd integer)
...
11 = 5 * 2 + 1 (=> 6th)
...
39 = 19 * 2 + 1 (=> 20th)

Sum from 1 to 11 = Sum of first 6th positive odd integers = \(6^2\) 

Sum from 1 to 39 = Sum of first 20th positive odd integers = \(20^2\) 

Sum of all odd integers between 13 and 39, inclusive
\(= 20^2 - 6^2\)
= (20-6) * (20+6)
= 14 * 26
= 364

 ­
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I think this is a high-quality question and I agree with explanation.
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Thanks good question.

Alternatively this can be solved in a more general approach using the Sum of an Arithmetic Sequence formula, shown below.

\(S_{n}=\frac{n}{2}[2a+(n-1)d]\)
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I like the solution - it’s helpful.
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I like the solution - it’s helpful.
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This is a great question that’s helpful for learning.
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I like the solution - it’s helpful.
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Easiest for me to solve this by recognizing that it is asking for sum of a consecutive set, which I know the formula for is:

Sum = Avg. * Number

Avg of consecutive set = (First + Last)/2
Number = (Last-first)/Interval + 1

Avg = 26
Number = 14

26*14 = 364
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I like the solution - it’s helpful. Sum = Avg x Quant

Avg = 39+13/(2) = 26
Quant = 39-13/2 (+1) = 14

26 x 14 = 364
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