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

∴ x^{2}-x+x+x^{2}=49x50

∴ x^{2} = 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.