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If the sum of the consecutive integers from –42 to n

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VP
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Re: If the sum of the consecutive integers from –42 to n  [#permalink]

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New post 13 Apr 2017, 18:49
this problem should be solved normally by using the following formula: k(k+1)/2
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Re: If the sum of the consecutive integers from –42 to n  [#permalink]

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New post 24 Jul 2018, 11:33
gmattokyo wrote:
If the sum of the consecutive integers from –42 to n inclusive is 372, what is the value of n?

A. 47
B. 48
C. 49
D. 50
E. 51


Using the equation below:

n*(n+1)/2 - 42*43/2 = 372

It is much easier to get n*(n+1) = 2550 => n=50
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If the sum of the consecutive integers from –42 to n  [#permalink]

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New post 24 Jul 2018, 21:58
gmattokyo wrote:
If the sum of the consecutive integers from –42 to n inclusive is 372, what is the value of n?

A. 47
B. 48
C. 49
D. 50
E. 51


sum of integers from -42 to 0 inclusive=(-42+0)/2*43=-903
372-(-903)=1275=sum of integers from 1 to n inclusive
1275=(n+1)/2*n→
n^2+n-2550=0→
(n+51)(n-50)=0
n=50
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Re: If the sum of the consecutive integers from –42 to n  [#permalink]

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New post 25 Jul 2019, 11:01
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Re: If the sum of the consecutive integers from –42 to n   [#permalink] 25 Jul 2019, 11:01

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