Question

The houses of a row are numbered consecutively from $1to49$. 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

Step 1: Find the sum of terms

First-term,$a=1$

The common difference, $d=1$

Let us say the number of $x^{th}$houses can be represented as;

Sum of nth term of AP$=\frac{n}{2}[2a+(n-1\left)d\right]$

Sum of number of houses beyond x house $=S_x-1$

$$\begin{gathered} =\frac{(x-1)}{2}[2.1+(x-1-1\left)1\right] \\ =\frac{(x-1)}{2}[2+x-2] \\ =\frac{x(x-1)}{2}\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \left(i\right) \end{gathered}$$

Step 2: Find the value of x

By the given condition, we can write,

$$\begin{gathered} S_{49}–S_x=\frac{49}{2}[2.1+(49-1\left)1\right]–\frac{x}{2}[2.1+(x-1\left)1\right] \\ =25\left(49\right)–\frac{x(x+1)}{2}\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots .\left(ii\right) \end{gathered}$$

As per the given condition, $eq.\left(i\right)andeq\left(ii\right)$are equal to each other;

Therefore,

$$\begin{gathered} \frac{x(x-1)}{2}=25\left(49\right)–\frac{x(x+1)}{2} \\ \frac{x(x-1)}{2}+\frac{x(x+1)}{2}=1225 \\ {x}^2=1225 \\ x=\pm 35 \end{gathered}$$

The number of houses cannot be a negative number.

Hence, the value of$x=35.$