Question 4

We need to find a house number $x$ where the sum of numbers before it equals the sum of numbers after it.

Step 1 — Set up the equation

The house numbers form an arithmetic progression. The first term $a$ is 1. The common difference $d$ is 1. The sum of the first $n$ terms is $S_n = \frac{n}{2}[2a + (n-1)d]$. The sum of houses preceding house $x$ is $S_{x-1}$. The sum of houses following house $x$ is $S_{49} - S_x$. We set these two sums equal to each other.

$S_{x-1} = S_{49} - S_x$

$\boxed{\text{Equation: } S_{x-1} = S_{49} - S_x}$

Step 2 — Substitute and solve for x

Let's substitute the formula for $S_n$ into our equation. We use $a = \mathbf{1}$ and $d = \mathbf{1}$.

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

Let's simplify the terms inside the brackets.

$\frac{x-1}{2}[x] = \frac{49}{2}[50] - \frac{x}{2}[x+1]$

Now, we multiply the entire equation by 2 to clear denominators.

$(x-1)x = 49(50) - x(x+1)$

Let's expand both sides of the equation.

$x^2 - x = 2450 - (x^2 + x)$

We distribute the negative sign on the right side.

$x^2 - x = 2450 - x^2 - x$

We add $x^2$ to both sides of the equation.

$2x^2 - x = 2450 - x$

Now, we add $x$ to both sides.

$2x^2 = 2450$

We divide both sides by 2.

$x^2 = \frac{2450}{2}$

$x^2 = 1225$

Finally, we take the square root of both sides.

$x = \sqrt{1225}$

$x = 35$

Since $x$ is a house number, it must be positive. The value of $x$ is 35.

Answer

The value of $x$ is 35.