Q4:

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 : $S_{x – 1} = S_{49} – S_x$]

Given: A row of houses numbered consecutively from $1, 2, 3, \dots, 49$. Let $x$ be the number of a specific house in this row.

To Find: The value of $x$ such that the sum of the house numbers preceding $x$ is equal to the sum of the house numbers following $x$.

Step 1: Defining the Sums

The sequence of house numbers is an Arithmetic Progression (AP) where the first term $a = 1$ and the common difference $d = 1$.

The sum of the first $n$ terms of an AP is given by the formula: $S_n = \frac{n}{2}(a + l)$, where $l$ is the last term.

The sum of all house numbers from $1$ to $49$ is: $S_{49} = \frac{49}{2}(1 + 49) = \frac{49}{2}(50) = 49 \times 25 = 1225$.

Step 2: Formulating the Equation

Let the sum of the houses preceding house $x$ be $S_{x-1}$. This is the sum of numbers from $1$ to $x-1$. $S_{x-1} = \frac{(x-1)}{2}(1 + x - 1) = \frac{(x-1)x}{2}$.

The sum of the houses following house $x$ is the total sum $S_{49}$ minus the sum of houses up to $x$. Sum following $x = S_{49} - S_x$. $S_x = \frac{x(x+1)}{2}$.

According to the problem, the sum of houses preceding $x$ equals the sum of houses following $x$: $S_{x-1} = S_{49} - S_x$.

Step 3: Algebraic Deduction

Substitute the expressions into the equation: $\frac{x(x-1)}{2} = 1225 - \frac{x(x+1)}{2}$.

Multiply the entire equation by $2$ to clear the denominators: $x(x-1) = 2450 - x(x+1)$.

Expand the terms: $x^2 - x = 2450 - (x^2 + x)$. $x^2 - x = 2450 - x^2 - x$.

Add $x$ to both sides: $x^2 = 2450 - x^2$.

Add $x^2$ to both sides: $2x^2 = 2450$.

Step 4: Solving for $x$

Divide by $2$: $x^2 = 1225$.

Take the square root of both sides: $x = \sqrt{1225}$. Since $x$ must be a positive integer representing a house number: $x = 35$.

Justification: Sum of houses preceding $35$ (i.e., $1$ to $34$): $S_{34} = \frac{34(35)}{2} = 17 \times 35 = 595$. Sum of houses following $35$ (i.e., $36$ to $49$): $S_{49} - S_{35} = 1225 - \frac{35(36)}{2} = 1225 - (35 \times 18) = 1225 - 630 = 595$. Since $595 = 595$, the condition is satisfied.

Final Answer: The value of $x$ is 35.