Question

The houses in a row are numbered consecutively from 1 to 49. Show that there exists a value of X such that sum of numbers of houses proceeding the house numbered X is equal to sum of the numbers of houses following X.

Answer 1

Given, the houses in a row numbered consecutively from 1 to 49.

Now, sum of numbers preceeding the number X

= $\frac{X(X-1)}{2}$

And, sum of numbers following the number X

= $\frac{49\left(50\right)}{2}-\frac{X(X-1)}{2}-X$

= $\frac{2450-X^2+X-2X}{2}$

= $\frac{2450-X^2-X}{2}$

According to the given condition,

Sum of no's precceding X = Sum of no's following X

$\frac{X(X-1)}{2}=\frac{2450-X^2-X}{2}$

$X^2-X=2450-X^2-X$

$2X^2=2450$

$X^2=1225$

$X=35$

Hence, at $X=35$, sum of no. of houses preceeding the house no. X is equal to sum of the no, of houses following X.