Question

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 $x$.

Answer 1

Here $a=1,d=1$ and $a_{49}=49$

As per question,

$S_{x-1}=S_{49}-S_x$............(i)$

$S_x=\frac{x}{2}[2+(x-1)\times 1]$

$=$ $\frac{x}{2}(x+1)=\frac{x^2+x}{2}$

Similarly,

$S_{x-1}=\frac{x-1}{2}[2+(x-1-1)\times 1]$

$=$ $\frac{x-1}{2}(2+x-2)=\frac{(x-1)x}{x}\frac{x^2-x}{2}$

Similarly,

$S_{49}=\frac{49}{2}[2+48\times 1]=\frac{49}{2}\times 50=1225$

After substituting the values of $S_{x-1},S_{49}$ and $S_x$ in equation (1), we get

$S_{x-1}==S_{49}-S_x$

$\frac{x^2-x}{2}=1225-\frac{x^2+x}{2}$

$\frac{x^2-x}{2}+\frac{x^2+x}{2}=1225$

$\frac{x^2-x+x^2-x}{2}=1225$

$x^2=1225$

$x=35$