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 sequence of houses is \(1, 2, 3, ..., 49\). This is an arithmetic progression (A.P.) with first term \(a = 1\) and common difference \(d = 1\). Let the \(x^{th}\) house be considered.

The sum of the first \(n\) terms of an A.P. is given by the formula: Sum = \(\frac n2 [2𝑎+(𝑛−1)𝑑]\).

The sum of the numbers of houses preceding the \(x^{th}\) house is \(S_{x-1}\):

\(S_{x-1} = \frac{(x-1)}{2}[2a+(x-1-1)d]\)

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

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

\(= \frac{1}{2}x(x-1)\)

The sum of the numbers of houses following the \(x^{th}\) house is \(S_{49} - S_x\):

\(S_{49} - S_x = \frac{49}{2}[2(1)+(49-1)1] - \frac{x}{2}[2(1)+(x-1)1]\)

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

\(= \frac{49}{2} \times 50 - \frac{1}{2}x(x+1)\)

It is given that these two sums are equal.

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

\(\frac{x^2}{2} - \frac{x}{2} = 1225 - \frac{x^2}{2}-\frac{x}{2}\)

\(x^2 = 1225\)

\(x = ± 35\)

Since house numbers must be positive integers, \(x = 35\).

Therefore, the \(35^{th}\) house is such that the sum of the numbers of houses preceding it is equal to the sum of the numbers of houses following it.