How to approach the attached question? In under 2 mins!!!
OA to be discussed!
Sum of K positive integers
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The sum of the first k positive integers is k(k+1)/2
The sum of the n integers is n(n+1)/2
The sum of the m integers is, m(m+1)/2
Since, 0<n<m, The sum of the integers from n to m is, m(m+1)/2 - n(n+1)/2
IMO B.
The sum of the n integers is n(n+1)/2
The sum of the m integers is, m(m+1)/2
Since, 0<n<m, The sum of the integers from n to m is, m(m+1)/2 - n(n+1)/2
IMO B.
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Since 0 < n < m
the sum of all integers from n to m (both inclusive) will be sum of first m integers - the sum if (n-1) integers.
If you plug in these into the formula...you would get
[m(m+1)/2] - [(n-1)n/2]
This will give us the answer C
Now to verify this lets say n = 3 and m = 5.
The answer should be 3 + 4 + 5 = 12
if we use the method above
A )sum of m numbers = 5 + 4 + 3 + 2 + 1 = 15.
B) sum of n-1 numbers = 2 + 1 = 3
Hence A - B = 15 - 3 = 12 (Bingo!!)
the sum of all integers from n to m (both inclusive) will be sum of first m integers - the sum if (n-1) integers.
If you plug in these into the formula...you would get
[m(m+1)/2] - [(n-1)n/2]
This will give us the answer C
Now to verify this lets say n = 3 and m = 5.
The answer should be 3 + 4 + 5 = 12
if we use the method above
A )sum of m numbers = 5 + 4 + 3 + 2 + 1 = 15.
B) sum of n-1 numbers = 2 + 1 = 3
Hence A - B = 15 - 3 = 12 (Bingo!!)
fruti_yum wrote:How to approach the attached question? In under 2 mins!!!
OA to be discussed!
Attempt 1: 710, 92% (Q 42, 63%; V 44, 97%)
Attempt 2: Coming soon!
Attempt 2: Coming soon!