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.

1, 2, 3, ….., x, ……., 49.

The above list is an A.P.,

First term (a) = 1 and common difference (d) = 2 - 1 = 1.

By formula,

Sum of n terms = S = $\dfrac{n}{2}[2a + (n - 1)d]$

Given,

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.

∴ S_{x - 1} = S₄₉ - S_x

Substituting values we get :

$\Rightarrow \dfrac{x - 1}{2}[2 \times 1 + (x - 1 - 1) \times 1] = \dfrac{49}{2}[2 \times 1 + (49 - 1) \times 1] - \dfrac{x}{2}[2 \times 1 + (x - 1) \times 1] \\[1em] \Rightarrow \dfrac{x - 1}{2}[2 + x - 2] = \dfrac{49}{2}[2 + 48] - \dfrac{x}{2}[2 + x - 1] \\[1em] \Rightarrow (x - 1)x = 49 \times 50 - x(x + 1) \\[1em] \Rightarrow x^2 - x = 2450 - x^2 - x \\[1em] \Rightarrow x^2 + x^2 - x + x = 2450 \\[1em] \Rightarrow 2x^2 = 2450 \\[1em] \Rightarrow x^2 = 1225 \\[1em] \Rightarrow x = \sqrt{1225} = 35.$

Hence, x = 35.