Houses
The houses of a row are numbered consecutively from 1 to 49. Find the value of such that the sum of the numbers of houses preceding the house numbered is equal to the sum of the numbers of the houses following it.
The answer is 35.
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1 solution
let n = number we're looking for.
1 to n - 1 is the sequence of numbers before n
n+1 to 49 is the sequence of numbers after n
Using the summation formula for an arithmetic series
SUM = number / 2 * ( first + last)
Sum ( 1 to n -1 ) = (n-1)/2 * ( 1 + n -1) = (n-1)*(n) / 2 = (n^2 - n) / 2
Sum (n + 1 to 49 ) = (49 - n) / 2 * (n + 1 + 49) = (49 - n) * (50 + n) / 2
These two sums must be equal so since denominators are equal the numerators are equal:
Multiplying second sum to remove ( )
n^2 - n = 2450 - n -n^2
2*n^2 = 2450
n^2 = 1225
n = 35