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 number of the houses preceeding the house number x is equal to the sum of the number of the houses following it. Find the value of x.

Answer 1

Let there be a value of x such that the sum of the number of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it

House $H_1,H_2,H_3.......H_{x-1},H_{x+1}........H_{49}$
House No. 1 2 3 .......... x -1 x + 1..............49

House number will form an A.P whose first term and the common difference is 1

Sum of n terms $S=\frac{n}{2}[2a+(n-1)\times d]$

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

$\Rightarrow \frac{(x-1)}{2}\left[2\right(1)+(x-1-1\left)1\right]=\frac{49}{2}[2+48]-\frac{x}{2}\left[2\right(1)+(x-1\left)1\right]$

$\Rightarrow \frac{x-1}{2}[2+(x-2\left)\right]=\frac{49}{2}\left[50\right]-\frac{x}{2}[2+x-1]$

$\Rightarrow \frac{x-1}{2}\left[x\right]=\frac{49}{2}\left[50\right]-\frac{x}{2}[x+1]$

$\Rightarrow \frac{x}{2}[x-1+x+1]=\frac{49}{2}\left[50\right]$

$\Rightarrow \frac{x}{2}\left[2x\right]=49\times 25$

$\Rightarrow x^2=49\times 25$

$\Rightarrow x=7\times 5=35.$

Since $x$ is not a fraction hence the value of x satisfying the given condition exists and is equal to $35$.