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The sum of the consecutive integers from 2 to 13 [#permalink]
20 Oct 2017, 02:50
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Question Stats:
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40% (00:49) wrong based on 119 sessions
Quantity A 
Quantity B 
The sum of the consecutive integers from 2 to 15 
34 less than the sum of the consecutive integers from 1 to 17 
A) Quantity A is greater. B) Quantity B is greater. C) The two quantities are equal. D) The relationship cannot be determined from the information given.
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Re: The sum of the consecutive integers from 2 to 13 [#permalink]
20 Oct 2017, 04:46
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Carcass wrote:
Quantity A 
Quantity B 
The sum of the consecutive integers from 2 to 13 
34 less than the sum of the consecutive integers from l to 17 
A) Quantity A is greater. B) Quantity B is greater. C) The two quantities are equal. D) The relationship cannot be determined from the information given. What's l in quantity B? Is it 1?



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Re: The sum of the consecutive integers from 2 to 13 [#permalink]
03 Mar 2018, 12:04
I think the answer is given wrong, for the QA 132=11 then 11/2= 5.5, so 5.5 * 15 (pair) = 82.5 for QB 1+17=18, then 171=16 (because no mentioning about inclusiveness), so 16/2=8 (pair) and 8*18=144, 14436=108, B>A
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Re: The sum of the consecutive integers from 2 to 13 [#permalink]
03 Mar 2018, 19:28
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Is the answer C correct here ?



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Re: The sum of the consecutive integers from 2 to 13 [#permalink]
05 Mar 2018, 14:12
Yes, it is C. I give you kudos for the right answer but next time give also us your reasoning. Regards
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Re: The sum of the consecutive integers from 2 to 13 [#permalink]
05 Mar 2018, 16:56
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I dont see how the answer can be C.
I used Gauss equation to solve this and I even counted all numbers from 1 to 17 and from 2 to 13.
Gauss euqation:
S= (N(A + z))/2
S = sum N = number of terms in the set A = First number in the set Z = final number in the set
Quantity A
s = ((12)(2 + 13))/2 s = 90
Quantity A = 90
Qunatity B
s = (17(1 + 17))/2 s = 153
153  34 = 119
quantity B = 119
A<B
Answer B
Am I missing something obvious here?



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Re: The sum of the consecutive integers from 2 to 13 [#permalink]
05 Mar 2018, 18:00
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Answer B Some of the consecutive integers equal: (first number + last number) * [(number of integers) / 2] and why is that? considering the sequence 1, 2, 3, 4, 5, 6 their sum is (1+6) + (2+5) + (3+4) = 7 + 7 + 7 = 7 * 6/2 = 7 *3 = 21 a1 + a2 + a3 + .... + an2 + an1 + an = (an+a1) + (an1 + a2) +....= n/2 * (an+a1)
so for the integers between 2 and 13 sum is: (2+13) * (132+1)/2 = 90. A equals 90. and for integers between 1 and 17 sum is: (1+17) * 17/2 = 153 and B equals 153  34 = 119 so B is bigger than A.
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Re: The sum of the consecutive integers from 2 to 13 [#permalink]
05 Mar 2018, 20:32
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(The problem has been edited since this response. In the original problem, Quantity A was the sum of integers from 2 to 13.) There is definitely something wrong with this problem. Assuming l in quantity B means 1, the answer is given as C, but it should be B. Let's prove it. As several people have shown, you can use the average formula, or Sum/(# of items) = Ave, to find the totals of both sides. I would do it more simply in this case, without the formula. There is a great deal of overlap between the two sides. After all, the integers 2 through 13 are contained inside the integers from 1 through 17. Let's subtract all numbers that are in both sets from both sides. So what numbers are left over? Quantity A is totally contained within Quantity B, so at this point it's got nothing left and is 0. What about Quantity B? Quantity B has 1, 14, 15, 16, and 17, while Quantity A does not. Adding these we get 63. (A quick way to add the last four numbers would be to add 14 and 17 to get 31, and double that since 15 + 16 must be the same, to get 62, and then adding 1.) Subtracting 34 from 63 will clearly get us something bigger than 0, so the answer should be B, not C. BONUS PROBLEM: If the answer were legitimately C, what would l have to be? The only way C could be correct is if l were not 1, but some other integer. If we set the two quantities equal to each other we will get: 90 = ((l + 17)/2)(17  l + 1)  34 The two parenthesis on the right represent the average of the integers and the number of integers, respectively. Next, we have: 124 = .5(l + 17)(18  l) 248 = (l + 17)(18  l) This will be a quadratic equation: 248 = 18l  l^2 + 17x18  17l 248 = l  l^2 + 306 l^2  l  58 = 0 This quadratic can't be factored with integers. 8 comes closest but it doesn't quite work. So basically this proves that there is no way for the two quantities to be equal. There's probably a typo somewhere, even besides the l.
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Last edited by SherpaPrep on 06 Mar 2018, 08:39, edited 1 time in total.



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Re: The sum of the consecutive integers from 2 to 13 [#permalink]
06 Mar 2018, 07:13
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Sorry guys. Thanks for your precious replies. there was a typo in the first quantity. The sum of the consecutive integers from 2 to 15Thank you so much. Regards
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Re: The sum of the consecutive integers from 2 to 13 [#permalink]
06 Mar 2018, 09:51
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There are several ways to go about this problem. (The new, edited one.) You can actually add up all the integers in both sides using some variation of the average formula, but this seems needlessly complicated. We can clearly see that Quantity B is mostly overlapped with Quantity A. If we subtract out every integer on both sides, we will vastly simplify the problem. Subtracting the integers from 2 to 15 on both sides leaves us with 0 under Quantity A and under Quantity B, we're left with 1, 16, and 17. These add up to 34 and since we must subtract 34 from Quantity B, we'll wind up with 0 on both sides. Thus the answer is C.
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Re: The sum of the consecutive integers from 2 to 13 [#permalink]
07 Jun 2018, 22:08
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Answer is C.
Let's take option A: Sum of consecutive integers from 2>15 is = Sum of consecutive integers from 1>15 1 i.e (15)(15+1)/2 1 => 119
Now option B: Sum of consecutive integers from 1>17 is =(17)(17+1)/2 => 153, 15334 =>119
Thus, C



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Re: The sum of the consecutive integers from 2 to 13 [#permalink]
25 Jun 2018, 21:06
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so first, in option B, it is 1 to 17. About solving the problem, we do not want to waste much time on test day and also get the right answer. So think this way both A and B have a set in common that is 2 to 15. Now, the remaining number in B are 1+16+17 which equals 34. So when we subtract 34 from sum of 1 to 17, as given, we will get the same answer as sum of 2 to 15.



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Re: The sum of the consecutive integers from 2 to 13 [#permalink]
09 Jul 2018, 00:50
Carcass wrote: Sorry guys. Thanks for your precious replies.
there was a typo in the first quantity.
The sum of the consecutive integers from 2 to 15
Thank you so much.
Regards According to the rule: sum of consecutive integers is a1+an/2 * n. So quantity A will be as follows: Quant A: (2+15)/2 * 14 = 119 not 119. Quant B: (1+17)/2 * 17 = 153  35 = 119. Answer should not be equal?
Last edited by Avraheem on 14 Jul 2018, 00:51, edited 1 time in total.



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Re: The sum of the consecutive integers from 2 to 13 [#permalink]
09 Jul 2018, 02:11
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Apply the formula for sum of a arithmetic progression ( https://www.wikiwand.com/en/Arithmetic_progression#/Sum) QA = (2+15)/2*(122+1) = 109 = QB = (17+1)/2*1734 = 109 > C.
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Re: The sum of the consecutive integers from 2 to 13 [#permalink]
09 Jul 2018, 08:15
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Quote:
Quantity A 
Quantity B 
The sum of the consecutive integers from 2 to 15 
34 less than the sum of the consecutive integers from l to 17 
A) Quantity A is greater. B) Quantity B is greater. C) The two quantities are equal. D) The relationship cannot be determined from the information given. B = [ 1 + (the sum of the consecutive integers from 2 to 15) + 16 + 17]  34. Since the values in red all cancel out, we get: B = the sum of the consecutive integers from 2 to 15. Thus, A and B are equal.
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Re: The sum of the consecutive integers from 2 to 13
[#permalink]
09 Jul 2018, 08:15





