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Difficulty: 505-555 Levelx   Arithmeticx                           
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Re: The sum of all the integers k such that −26 < k < 24 is [#permalink]
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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: \(\frac{-25+23}{2}*49=-49\).

Answer: D.

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

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


-25, -24, -23 .... 23
-23 to 23 cancel each other.

Remaining: -25, -24

Sum = -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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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 number of integers is from -25 to 23 inclusive \(= 23-(-25)+1=23+25+1=49\)

The sum \(= \frac{23+(-25)*49}{2}=\frac{-2*49}{2}=-49\)

The answer is \(D\)
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Re: The sum of all the integers k such that −26 < k < 24 is [#permalink]
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can someone explain me why is -25 the largest number and 23 the smallest number in the average approach. Thank you in advance.
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The sum of all the integers k such that −26 < k < 24 is [#permalink]
JCGF2021 wrote:
can someone explain me why is -25 the largest number and 23 the smallest number in the average approach. Thank you in advance.


Hi, the reason why -25 is the larger number is because k > -26. What is the next integer that is greater than -26?
Similarly, the smaller number is 23 because k < 24. What is the integer that is immediately before 24?
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Re: The sum of all the integers k such that −26 < k < 24 is [#permalink]
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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.


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Answer: Option D

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Re: The sum of all the integers k such that −26 < k < 24 is [#permalink]
Correct Option : D (-49)
-26<K<24
K range is greater than -26 i.e -25 and less than 24 i.e 23
{-23 to 23} addition is zero, remains with {-25, -24} i.e (-25)+(-24) = -49
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Re: The sum of all the integers k such that 26 < k < 24 is [#permalink]
Hi MHIKER, why are you adding 1 to get 49? Thanks
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