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The 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 [#permalink]

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20 Oct 2015, 07:57

2

This post received KUDOS

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.

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

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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.

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 evenly-spaced 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 [#permalink]

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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

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: \(\frac{-25+23}{2}*49=-49\).

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

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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+(n-1)*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

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

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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.

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

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12 Nov 2017, 03:51

ScottTargetTestPrep wrote:

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 evenly-spaced 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

Hello, what if we didn't have evenly spaced integers, how would we find average?