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

Given:

The houses of a row are numbered consecutively from 1 to 49.

To do:

We have to show that there is 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 and find this value of $x$.

Solution:

Let there be 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.

This implies,

$1\ +\ 2\ +\ 3\ +\ .\ .\ .\ .\ +\ ( x\ –\ 1)=( x+1)+( x+2)+\ .\ .\ .\ .\ .\ + 49$

$\therefore 1+2+3+\ .\ .\ .\ .\ +( x–1)=[1+2+\ ....+x+( x–1)+....+49] \ –\ ( 1+2+3+.\ .\ .\ .\ +x$

Sum of the $n$ terms of the A.P. $S_{n} =\frac{n}{2}( a+l)$   [a is the first term and l is the last term of the given A.P.]

$\frac{x-1}{2}( 1+x-1) =\frac{49}{2}( 1+49) -\frac{x}{2}( a+l)$

$\therefore  x( x\ –\ 1) \ =\ 49\ \times \ 50\ –\ x( 1\ +\ x)$

$\therefore  x( x\ –\ 1) \ +\ x( 1\ +\ x) \ =\ 49\ \times \ 50$

$\therefore  x( x-1+1+x) =49\times 50$

$2x^{2} = 49\times50$

$\therefore x^{2} =49\times 25$

$\Rightarrow x=\sqrt{49\times 25}$

$\therefore x=7\times 5=35$

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