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

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

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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  [#permalink]

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25 Oct 2016, 17:05
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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.

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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
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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.

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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, 05:31
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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
the-sum-of-all-the-integers-k-such-that-26-k-24-is-72685.html

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

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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, 09:15
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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  [#permalink]

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20 Oct 2015, 09:42
2
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 0-23.

We can conclude that for all but -25 and -24 the number pairs will add to 0. So we have left -25 - 24 = -49.

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

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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  [#permalink]

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24 Oct 2016, 10:59
1
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...

Answer will be (D) - 49
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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
2
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  [#permalink]

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13 Apr 2017, 12:09
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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: $$\frac{-25+23}{2}*49=-49$$.

Hope it's clear.
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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
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Re: The sum of all the integers k such that −26 < k < 24 is  [#permalink]

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02 Jun 2017, 09:18
1
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  [#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.
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Re: The sum of all the integers k such that −26 < k < 24 is  [#permalink]

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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  [#permalink]

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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 (b-a)-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  [#permalink]

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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  [#permalink]

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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.

-25-24-23-22-21........-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=[23-25/2]*[23-(-25)+1]= -1*49=-49
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The sum of all the integers k such that −26 < k < 24 is  [#permalink]

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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

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The sum of all the integers k such that −26 < k < 24 is   [#permalink] 11 May 2019, 22:42
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