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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