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.

Hint:

\(S_x – 1 = S_{49} – S_x\)

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 x^th 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 x^th 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 x^th 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.