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

The number of houses was
$1, 2, 3 ...... 49$
It can be observed that the number of houses are in an A.P. having a as 1 and d also as 1.
Let us assume that the number of xth house was like this.
We know that,
Sum of n terms in an A.P. =$\frac n2 [2𝑎+(𝑛−1)𝑑]$
Sum of number of houses preceding xth house = $S_x − 1$
= $\frac {(x-1)}{2}[2a+(x-1-1)d]$

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

= $\frac {x-1}{2}[2+x-2]$

= $\frac 12 x(x-1)$
Sum of number of houses following xth house =$S_{49} − S_{x}$

= $\frac {49}{2}[2(1)+(49-1)1] - \frac x2[2(1)+(x-1)1]$

= $\frac {49}{2}(2+49-1) -\frac x2(2+x-1)$

= $\frac {49}{2} \times 50 - \frac 12x(x+1)$
It is given that these sums are equal to each other.

$\frac {x(x-1)}{2}= 25 \times 49 - \frac 12x(x+1)$

$\frac {x^2}{2} - \frac x2 = 1225 - \frac {x^2}{2}-\frac x2$
$x^2 = 1225$
$x = ± 35$
However, the house numbers are positive integers.
The value of x will be $35$ only.

Therefore, house number $35$ is such that the sum of the numbers of houses preceding the house numbered $35$ is equal to the sum of the numbers of the houses following it.