Question

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> 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 $x.$
[ Hint : $S_{x-1}=S_{49}-S_x$]

Answer 1

Row houses are numbers from $1,2,3,4,\mathrm{5...........49.}$
Thus we can see the houses numbered in a row are in the form of $AP.$
So,
First term, $a=1$
Let us say the number of $x^{th}$ house can be represented as ;
Sum of nth term of $AP=n/2[2a+(n-1\left)d\right]$
Sum of number of houses beyond $x$ house $=S_{x-1}$
$=(x-1)/2[2.1+(x-1-1\left)1\right]$
$=(x-1)/2[2+x-2]$
$=x(x-1)/2....................\left(i\right)$
By the given condition we can write,
$S_{49}-S_x=\left\{49/2\right[2.1+(49-1)1\left]\right\}-\left\{x/2\right[2.1+(x-1)1\left]\right\}$
$=25\left(49\right)-x(x+1)/2..................\left(ii\right)$

As per the given condition, eq (i) and eq(ii) are equal to each other,
Therefore,
$x(x-1)/2=25\left(49\right)-x(x-1)/2$
$x=\pm 35$
As we know, the number of houses cannot be a negative number Hence, the value of $x$ is $35$.