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.
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