question_answer

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:

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

Now, sum of numbers preceding the number X

\[=\frac{X(X-1)}{2}\]

And, sum of numbers following the number X

\[=\frac{49(50)}{2}-\frac{X(X-1)}{2}-X\]

\[=\frac{2450-{{X}^{2}}+X-2X}{2}\]

\[=\frac{2450-{{X}^{2}}-X}{2}\]

According to the given condition,

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

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

\[{{X}^{2}}-X=2450-{{X}^{2}}-X\]

\[2{{X}^{2}}=2450\]

\[{{X}^{2}}=1225\]

\[X=35\]

Hence, at \[X=35\], sum of no. of houses preceding the house no. X is equal to sum of the no. of houses following X.