The houses of a row are numbered consecutively from $1$ to $49$. Show that there is 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. Find this value of $x$.

$S_{x-1}=S_{49}-S_x$............(i)S_{x}=\dfrac{x}{2}\left [ 2+(x-1)\times 1 \right ]=\dfrac{x}{2}(x+1)=\dfrac{x^{2}+x}{2}$Similarly,$S_{x-1}=\dfrac{x-1}{2}\left [ 2+(x-1-1)\times 1 \right ]=\dfrac{x-1}{2}(2+x-2)=\dfrac{(x-1)x}{x}\dfrac{x^{2}-x}{2}$Similarly,$S_{49}=\dfrac{49}{2}\left[ 2+48\times 1 \right ]=\dfrac{49}{2}\times 50=1225$Aftersubstitutingthevaluesof$S_{x-1}, S_{49}$and$S_x$inequation(1),weget$S_{x-1}==S_{49}-S_{x}\dfrac{x^{2}-x}{2}=1225-\dfrac{x^{2}+x}{2}\dfrac{x^{2}-x}{2}+\dfrac{x^{2}+x}{2}=1225\dfrac{x^{2}-x+x^{2}-x}{2}=1225x^{2}=1225x=35$$