The houses in a row numbered consecutively from 1 to 49. Show that there exists a value of X such that sum of numbers of houses preceding the house numbered X is equal to sum of the numbers of houses following X.
Solution
Let there be a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it.
That is, 1 + 2 + 3 . . . . + (x -1) = (x + 1) + (x + 2) + . . . . . +49
∴ 1 + 2 + 3 + . . . . + (x - 1)
= [1 + 2 + ...... + x + (x + 1) + ....+ 49] - (1 + 2 + 3 + . . . . + x)
∴ `(x-1)/2[1+x-1]=49/2[1+49]-x/2[1+x]`
∴ x(x - 1) = 49 x 50 - x(1 + x)
∴ x(x - 1) + x(1 + x) = 49 x 50
∴ x2-x+x+x2=49x50
∴ x2 = 49x25
∴ x = 7 x 5 = 35
Since x is not a fraction, the value of x satisfying the given condition exists and is equal to 35.
