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The sum of all the integers k such that −26 < k < 24 is
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20 Oct 2015, 02:58
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The sum of all the integers k such that −26 < k < 24 is (A) 0 (B) −2 (C) −25 (D) −49 (E) −51 Kudos for a correct solution.
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Re: The sum of all the integers k such that −26 < k < 24 is
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25 Oct 2016, 17:05
Bunuel wrote: The sum of all the integers k such that −26 < k < 24 is
(A) 0 (B) −2 (C) −25 (D) −49 (E) −51 We must determine the sum of the consecutive integers from 25 to 23, inclusive. To determine the sum we can use the formula sum = average x quantity. To determine quantity, the number of consecutive integers, we compute the following: quantity = largest number – smallest number + 1 quantity = 23 – (25) + 1 = 23 + 25 + 1 = 49 Next we must determine the average. Since we have a set of evenlyspaced integers we can determine the average using the formula: average = (largest number + smallest number)/2. average = (25 + 23)/2 = 2/2 = 1 Finally we can determine the sum: sum = quantity x average sum = 49 x 1 = 49. Alternate solution: We must determine the sum of the consecutive integers from 25 to 23 inclusive. However, if we add 23 and 23, the sum will be 0, and so will be the sum of 22 and 22, 21 and 21, and so on. Therefore, the sum of each of these pairs of numbers (as long as they are opposites) is 0. The only numbers left that are not paired with their opposites are 25, 24 and 0. So the sum of all the integers from 25 to 23, inclusive, is the same as the sum of 25, 24 and 0, which is (25) + (24) + 0 = 49. Answer: D
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Re: The sum of all the integers k such that −26 < k < 24 is
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20 Oct 2015, 07:57
The range of k is from 25 to +23. The range includes 23 pairs of opposite numbers which nullify each other and we are left with just 24 & 25, the sum of which is 49.
Answer choice D




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Re: The sum of all the integers k such that −26 < k < 24 is
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20 Oct 2015, 05:31
Bunuel wrote: The sum of all the integers k such that −26 < k < 24 is
(A) 0 (B) −2 (C) −25 (D) −49 (E) −51
Kudos for a correct solution. Bunuel: I guess I have seen this question on Forum thesumofalltheintegersksuchthat26k24is72685.htmlThe sum of all the integers k such that −26 < k < 24 = SUm of all Integers from 25 to 23 SUM = (25)+(24)++(23) = (25)+(24)+ (23)+(22)+(22)+(23) = (49)+ (0) = 49 Answer: option D
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Re: The sum of all the integers k such that −26 < k < 24 is
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20 Oct 2015, 09:15
Range is given as 26<k<24 => where 26 and 24 are excluded We can say that 23 to +23 in the range would be cancelled out.. R = {25,24,23 .................+22,+23} => This would give us 25  24 = 49.



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Re: The sum of all the integers k such that −26 < k < 24 is
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20 Oct 2015, 09:42
Bunuel wrote: The sum of all the integers k such that −26 < k < 24 is
(A) 0 (B) −2 (C) −25 (D) −49 (E) −51
Kudos for a correct solution. Since k defines a range between −26 < k < 24 we can set 0 as the reference point for the negative values and positive values. The negative values will range from 25 to 0 whereas the positive values will range from 023. We can conclude that for all but 25 and 24 the number pairs will add to 0. So we have left 25  24 = 49. Answer D.



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Re: The sum of all the integers k such that −26 < k < 24 is
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09 Oct 2016, 13:05
Bunuel wrote: The sum of all the integers k such that −26 < k < 24 is
(A) 0 (B) −2 (C) −25 (D) −49 (E) −51
Kudos for a correct solution. a little lengthy, but prevents counting errors: \(Sum = \frac{[25(23)+1]}{2} * (25  23) = \frac{49}{2} * (2) = 49\)
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Re: The sum of all the integers k such that −26 < k < 24 is
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24 Oct 2016, 10:59
Bunuel wrote: The sum of all the integers k such that −26 < k < 24 is
(A) 0 (B) −2 (C) −25 (D) −49 (E) −51
Kudos for a correct solution. −26 < k < 24 = 25 , 24 , 23........ k ...........23 Only 25 &  24 will remain , all gets cancelled...Hence answer will be 49 Answer will be (D)  49
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Re: The sum of all the integers k such that −26 < k < 24 is
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26 Oct 2016, 19:05
Question can be solved using sum/# = avg equation:
We know that the total number (#) is 23(25)+1 = 49 We also know that since 49 is odd we can pull the 25th number in the sequence and that will be the average
Thus, the equation becomes SUM/49 = 1 > Manipulating this you will find SUM = 49



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Re: The sum of all the integers k such that −26 < k < 24 is
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13 Apr 2017, 12:09
Bunuel wrote: The sum of all the integers k such that −26 < k < 24 is
(A) 0 (B) −2 (C) −25 (D) −49 (E) −51
Kudos for a correct solution. The sum will be: (25)+(24)+(23)+...+(1)+0+1+..+23 > the sum of pairs 23 and 23, 22 and 22 and so on is 0 and we are left only with 25+(24)=49. Or: as we have evenly spaced set: the sum will be average of the first and the last terms multiplied be the # of terms: \(\frac{25+23}{2}*49=49\). Answer: D. Hope it's clear.
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Re: The sum of all the integers k such that −26 < k < 24 is
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02 Jun 2017, 02:26
to find the sum of a series in AP, sum= [(1st term + last term)/2]* no. of terms
now, to find the last term of a series in AP: tn=an+(n1)*d where, an=1st term, n=no. of terms and d=common difference −26 < k < 24 in this case, tn=23 an=25 and d=1 hence n=49 sum= [(first term+last term)/2]*no. of terms=[(25+23)/2]49=49



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Re: The sum of all the integers k such that −26 < k < 24 is
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02 Jun 2017, 09:18
Bunuel wrote: The sum of all the integers k such that −26 < k < 24 is
(A) 0 (B) −2 (C) −25 (D) −49 (E) −51
Kudos for a correct solution. \(−26 < k < 24\) So, \(−26 < k < 24\) = \( −25,−24 , −23,−22, −21..................21, 22 , 23\) Thus, answer must be (D) −49
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Re: The sum of all the integers k such that −26 < k < 24 is
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14 Jun 2017, 07:30
For me I had difficulty translating the question into meaning. The way the question is phrased implies that k is the sum of a set of integers, the value of which falls in the range of integers between 26 to 24 not that k was the range of integers itself.



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Re: The sum of all the integers k such that −26 < k < 24 is
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13 Sep 2017, 09:43
Sum = [(First term + Last term)/2]*total number of terms Here first term is 25 Last term is 23 Total number of terms are 49



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Re: The sum of all the integers k such that −26 < k < 24 is
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06 Nov 2017, 01:21
Bunuel wrote: The sum of all the integers k such that −26 < k < 24 is
(A) 0 (B) −2 (C) −25 (D) −49 (E) −51
Kudos for a correct solution. solved using (ba)1: = (26  24)  1 = (50)1 = 49



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Re: The sum of all the integers k such that −26 < k < 24 is
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23 Apr 2018, 20:50
I can't stop laughing on seeing a silly mistake on question like this.



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The sum of all the integers k such that −26 < k < 24 is
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11 May 2019, 15:03
We must determine the sum of the consecutive integers from 25 to 1 and from 1 to 23, then we add them together.
2524232221........1 is 1( 25+24+23+22+21......2+1) or 1(1+2......+21+22+23+24+25)
Now,we have the famous formula for the sum of consecutives integers :n(n+1)/2
So, (1+2......+21+22+23+24+25)= 25(25+1)/2=325
1* (1+2......+21+22+23+24+25)= 1* [ 25(25+1)/2]= 325 (1)
The same for:
(1+2.......+20+21+22+23)= 23(23+1)/2=276 (2)
The sum of all the integers k such that −26 < k < 24 is: (1)+ (2)
1* [ 25(25+1)/2]+23(23+1)/2= 325+276= 49
Alternate solution:
Sum= Average*Number
average = (largest number + smallest number)/2. Number= largest number – smallest number + 1
Sum=[2325/2]*[23(25)+1]= 1*49=49



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The sum of all the integers k such that −26 < k < 24 is
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11 May 2019, 22:42
Bunuel wrote: The sum of all the integers k such that −26 < k < 24 is
(A) 0 (B) −2 (C) −25 (D) −49 (E) −51
Kudos for a correct solution. Here, Summation from 25 to +23 is as follows: From 23 to +23=0 24 & 25 are left So, the summation of (25)+(24)=49 Answer is D Posted from my mobile device



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Re: The sum of all the integers k such that −26 < k < 24 is
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15 Jul 2019, 01:43
26,25,24,23......0......23,24
So K should be between 24 to 26 it means we can't include this two number
So till 23+23= 0 all get cancel
Remaining 2425= 49 ( ANS)




Re: The sum of all the integers k such that −26 < k < 24 is
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