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The sum of first \(N\) consecutive odd integers is \(N^2\) . What is the sum of all odd integers between 13 and 39 inclusive?
(C) 2008 GMAT Club - m20#31
* 351 * 364 * 410 * 424 * 450
how about the way that is explained in the question:
The sum of consecutive odd integers from 1 to 39 = [(39+1)/2]^2 = 400 The sum of consecutive odd integers from 1 to 11 = [(11+1)/2]^2 = 36 The sum of consecutive odd integers from 13 to 39 = 400 - 36 = 364 _________________
The sum of first \(N\) consecutive odd integers is \(N^2\) . What is the sum of all odd integers between 13 and 39 inclusive?
(C) 2008 GMAT Club - m20#31
* 351 * 364 * 410 * 424 * 450
how about the way that is explained in the question:
The sum of consecutive odd integers from 1 to 39 = [(39+1)/2]^2 = 400 The sum of consecutive odd integers from 1 to 11 = [(11+1)/2]^2 = 36 The sum of consecutive odd integers from 13 to 39 = 400 - 36 = 364
Hi thank you, this is the OA However I used the same methodology, but I included 13 as in the sum of consecutive odd integers from 1 to 13 and subtracted that from 400, thereby getting 351 as my answer. Why do we only calculate the sum from 1 to 11? I know I am missing something, sorry if this is a stupid question!
The sum of first \(N\) consecutive odd integers is \(N^2\) . What is the sum of all odd integers between 13 and 39 inclusive?
(C) 2008 GMAT Club - m20#31
* 351 * 364 * 410 * 424 * 450
how about the way that is explained in the question:
The sum of consecutive odd integers from 1 to 39 = [(39+1)/2]^2 = 400 The sum of consecutive odd integers from 1 to 11 = [(11+1)/2]^2 = 36 The sum of consecutive odd integers from 13 to 39 = 400 - 36 = 364
Hi thank you, this is the OA However I used the same methodology, but I included 13 as in the sum of consecutive odd integers from 1 to 13 and subtracted that from 400, thereby getting 351 as my answer. Why do we only calculate the sum from 1 to 11? I know I am missing something, sorry if this is a stupid question!
"13 and 39 inclusive" so if you subtract 13 also then you are only finding the sum of all odd numbers from 15-39
The way this question is written makes a person use "their" way (meaning the test maker's way) and in reality, just going through add adding them all up doesn't take more than 60 seconds.
13 15 17 19 21 23 25 27 29 31 33 35 37 39
I approach adding these up as in pairs: 11 & 39, 13 & 37, 15 & 35, 17 & 33, 19 & 31, 21 & 29, 23 & 27. Each of these 6 pairs = 50, so 50 * 6 = 300. Then you have 39 and 25 left, that's 64, so 364.
Not the most "advanced" method, but it certainly won't take over 60 seconds to complete this way. _________________
------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.
can someone please explain me this line - 39=13+(n-1)2
where can i c a summary of ways to know the amount of N in series?
thanks guys
So the problem states that it is every odd number in between 13 and 39 inclusive. To find N integers you just count off how many are included in this series (14). The line in question just sets up an equation to easily solve for N. Subtract 13, then divide by two then add one and you get N=14. To check just count off. By and large it is better to memorize that equation since there are some instances were counting off via a check is too time consuming.
i dont understand the purpose of the first statement portion of the question. how is that relevant to what the question is asking?
The purpose of the first sentence is two fold; 1) save you time adding consecutive odd integers 2) test quantitative reasoning. That being said, it is not necessary to solve the problem, but it will save you a minute.
Here is how it breaks down: The sum of first N consecutive odd integers is N^2. There are 20 consecutive odd integers from 0-39. Therefore, The sum of the first 20 consecutive odd integers is 20^2 Expressable as, The sum of 1+2...39 = 400 or 20^2 = 400
What is the sum of all odd integers between 13 and 39 inclusive? What is the sum of 13+14+15...39? Since 400 = [1+2...11] + [13+15...39] We can reason that 364 = [6^2] - [20^2]
Re: Sum of Consecutive Integers [#permalink]
06 Jun 2012, 18:24
1
This post received KUDOS
manulath:
Fool proof 2-minute method, list the integers from 13 to 39 and sum them.
Better method Let's use the equation given, we are looking for:
13 + 15 + ... + 39 = (1+3+ ... + 39) - (1+3+ ... + 11) = sum of odd numbers from 1 to 39 less sum of odd numbers from 1 to 11
Sum of odd numbers from 1 to 39 = 20^2 - there are twenty odd numbers: (39-1)/2+1 = 20 - use the equation given, 20^2
Sum of odd numbers from 1 to 11 = 6^2 - there are six odd numbers (11-1)/2+1 = 6
20^2 - 6^2 = 364 B)
Robust method Question is for an arithmetic series, starting number = 13, constant difference = 2, number of items = 14 sum = 14*(13+(14-1)/2*2) = 14*13 = 364
gmatclubot
Re: Sum of Consecutive Integers
[#permalink]
06 Jun 2012, 18:24