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Updated on: 27 Sep 2017, 03:21
Originally posted by MathRevolution on 25 Jan 2016, 17:40.
Last edited by Bunuel on 27 Sep 2017, 03:21, edited 3 times in total.
Last edited by Bunuel on 27 Sep 2017, 03:21, edited 3 times in total.
Added the OA.
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
52% (01:42) correct
48%
(01:32)
wrong
based on 135
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If the sum of the first k positive integers is equal to, k(k+1)/2, What is the sum of the integers from n to m, inclusive, where 0<n<m?
A. (m(m+1)-(n-1)n)/2
B. (m(m+1)-(n+1)n)/2
C. (m(m-1)-(n-1)n)/2
D. (m(m-1)-(n+1)n)/2
E. (m(m+1)-(n-1)n)/4
* A solution will be posted in two days.
OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/the-sum-of-t ... 26289.html
A. (m(m+1)-(n-1)n)/2
B. (m(m+1)-(n+1)n)/2
C. (m(m-1)-(n-1)n)/2
D. (m(m-1)-(n+1)n)/2
E. (m(m+1)-(n-1)n)/4
* A solution will be posted in two days.
OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/the-sum-of-t ... 26289.html
26 Jan 2016, 01:06
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MathRevolution
1+2+3+4..+n+...+m
m(m+1)/2 - (n-1)(n-1+1)/2
= (m(m+1) - (n-1)(n))/2
Answer A
27 Jan 2016, 18:23
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If the sum of the first k positive integers is equal to, k(k+1)/2, What is the sum of the integers from n to m, inclusive, where 0<n<m?
A. (m(m+1)-(n-1)n)/2
B. (m(m+1)-(n+1)n)/2
C. (m(m-1)-(n-1)n)/2
D. (m(m-1)-(n+1)n)/2
E. (m(m+1)-(n-1)n)/4
==> n+(n+1)+…….+(m-1)+m={1+2+…+(n-1)+n+(n+1)+…+(m-1)+m}-{1+2+…+(n-1)}=m(m+1)/2-n(n-1)/2.
Therefore, the answer is A.
A. (m(m+1)-(n-1)n)/2
B. (m(m+1)-(n+1)n)/2
C. (m(m-1)-(n-1)n)/2
D. (m(m-1)-(n+1)n)/2
E. (m(m+1)-(n-1)n)/4
==> n+(n+1)+…….+(m-1)+m={1+2+…+(n-1)+n+(n+1)+…+(m-1)+m}-{1+2+…+(n-1)}=m(m+1)/2-n(n-1)/2.
Therefore, the answer is A.
