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Find m if the sum of the consecutive integers from -16 to m, inclusive

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Find m if the sum of the consecutive integers from -16 to m, inclusive  [#permalink]

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New post Updated on: 09 Jun 2017, 04:14
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Find m if the sum of the consecutive integers from -16 to m, inclusive is 74.


A) 17
B) 18
C) 19
D) 20
E) 21

Originally posted by Mina Satti on 09 Jun 2017, 00:02.
Last edited by Bunuel on 09 Jun 2017, 04:14, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Find m if the sum of the consecutive integers from -16 to m, inclusive  [#permalink]

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New post 09 Jun 2017, 00:06
1
It will be D 20
As -16 to 16 will cancel out the addition. And then you can simply add 17 ,18 ,19 and 20 to get 74.

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Re: Find m if the sum of the consecutive integers from -16 to m, inclusive  [#permalink]

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New post 09 Jun 2017, 03:24
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Mina Satti wrote:
Question. Find m if the sum of the consecutive integers from -16 to m, inclusive is 74.


A) 17, B)18 C) 19 D) 20


Sum of all consecutive integers from -16 to +16 = 0

Now sum of 17+18+19+20=74

Hence m= 20

Answer option D

Bunuel please increase one option here to make it a GMAT like question .

Add an option E)21
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Re: Find m if the sum of the consecutive integers from -16 to m, inclusive  [#permalink]

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New post 09 Jun 2017, 04:16
GMATinsight wrote:
Mina Satti wrote:
Question. Find m if the sum of the consecutive integers from -16 to m, inclusive is 74.


A) 17, B)18 C) 19 D) 20


Sum of all consecutive integers from -16 to +16 = 0

Now sum of 17+18+19+20=74

Hence m= 20

Answer option D

Bunuel please increase one option here to make it a GMAT like question .

Add an option E)21


Done. Thank you for the suggestion.
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Re: Find m if the sum of the consecutive integers from -16 to m, inclusive  [#permalink]

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New post 09 Jun 2017, 06:52
Imo D
Can be calculated from sum formula of a sequence

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Re: Find m if the sum of the consecutive integers from -16 to m, inclusive  [#permalink]

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New post 09 Jun 2017, 12:01
Opt D
Other way round
S = n/2(2a+(n-1)d)
=> 74=n/2(-32+n-1) as common difference is 1
148=-33n+n^2
n^2-33n-148=0
upon simplification n = 37
tn=a+(n-1)d
m=16+(37-1)1
m=20
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Re: Find m if the sum of the consecutive integers from -16 to m, inclusive  [#permalink]

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New post 13 Jun 2017, 11:08
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Mina Satti wrote:
Find m if the sum of the consecutive integers from -16 to m, inclusive is 74.


A) 17
B) 18
C) 19
D) 20
E) 21


We can use the following formula:

sum = average x quantity

Since we have a set of evenly spaced integers, the average is (-16 + m)/2.

We can also calculate the quantity as m - (-16) + 1 = m + 17.

Thus:

74 = (-16 + m)/2 x (m + 17)

148 = (m - 16)(m + 17)

148 = m^2 + m - 272

m^2 + m - 420 = 0

(m + 21)(m - 20) = 0

m = -21 or m = 20

Since m > -16, m must be 20.

Alternate Solution:

The integers in the sum will be -16, -15, -14, …, -2, -1, 0, +1, +2, …, m. Notice that all the negative integers from -16 to -1 inclusive will be offset by the positive integers from +1 to +16, yielding a net sum of 0. So now, starting with the integer +17, we need to add consecutive integers until we get a sum of 74. We see that 17 + 18 + 19 + 20 = 74, and thus m = 20, the largest integer in the set.

Answer: D
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Find m if the sum of the consecutive integers from -16 to m, inclusive  [#permalink]

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New post 12 Jan 2019, 20:26
Mina Satti wrote:
Find m if the sum of the consecutive integers from -16 to m, inclusive is 74.


A) 17
B) 18
C) 19
D) 20
E) 21



[(-16+m)/2](17+m)=74→
m^2+m-420=0→
x=20
D
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Find m if the sum of the consecutive integers from -16 to m, inclusive &nbs [#permalink] 12 Jan 2019, 20:26
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Find m if the sum of the consecutive integers from -16 to m, inclusive

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