# Ex.5.4 Q4 Arithmetic progressions Solutions - NCERT Maths Class 10

## Question

The houses of a row are number consecutively from \(1\) to \(49.\) Show that there is a value of *\(x\) *such that the sum of numbers of the houses preceding the house numbered *\(x\)* is equal to the sum of the number of houses following it. Find this value of \(x.\)

[Hint: \({S_x}_{ - 1} = {S_{49}} - {S_x}\)]

## Text Solution

**What is Known?**

The houses of a row are number consecutively from \(1\) to \(49\)

**What is Unknown?**

Value of *\(x\)*

**Reasoning:**

Sum of the first \(n\) terms of an AP is given by \({S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\)

Where \(a\) is the first term, \(d\) is the common difference and \(n\) is the number of terms.

**Steps:**

Given:

Number of houses are \(1,2,3, .................49\)

By Observation, the numbers of houses are in an A.P.

Hence

- First term, \(a = 1\)
- Common difference, \(d = 1\)

Let us assume that the number of *\(x^\rm{th}\)* house can be expressed as below:

We know that ,

Sum of n terms in an A.P. \({S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\)

Sum of numbers of houses preceding \(x^\rm{th}\) house \(= S_{x−1}\)

\[\begin{align}{S_{x - 1}} &\!=\!\frac{{\left( {x - 1} \right)}}{2}\left[\!{2a \!+\!\left( {\left( {x\!-\!1} \right) \!- 1} \right)d} \right]\\& = \frac{{\left( {x\!-\!1} \right)}}{2}\left[ {2\!\times\! 1\!+\!\left( {x\!-\!2} \right) \!\times\! 1} \right]\\ &= \frac{{\left( {x - 1} \right)}}{2}\left[ {2 + x - 2} \right]\\ &= \frac{{x\left( {x - 1} \right)}}{2}\end{align}\]

By the given we know that, Sum of number of houses following \(x^\rm{th}\) house \(= S_{49} – S_{x}\)

\[\begin{align} {{S}_{49}}-{{S}_{x}}&=\left\{ \begin{array} & \frac{49}{2}\left[ 2\times 1+\left( 49-1 \right)\times 1 \right] \\ - \frac{x}{2}\left[ 2\times 1+\left( x-1 \right)\times 1 \right] \\ \end{array} \right\} \\ & =\frac{49}{2}\left[ 2+48 \right]-\frac{x}{2}\left[ 2+x-1 \right] \\ & =\frac{49}{2}\left[ 2+48 \right]-\frac{x}{2}\left[ 2+x-1 \right] \\ & =\frac{49}{2}\times 50-\frac{x}{2}\left[ x+1 \right] \\ & =1225-\frac{x\left( x+1 \right)}{2} \\ \end{align}\]

It is given that these sums are equal.

\[\begin{align}\frac{{x\left( {x - 1} \right)}}{2} &= 1225 - \frac{{x\left( {x + 1} \right)}}{2}\\\frac{{{x^2}}}{2} - \frac{x}{2} &= 1225 - \frac{{{x^2}}}{2} - \frac{x}{2}\\{x^2} &= 1225\\x &= \pm 35\end{align}\]

As the number of houses cannot be a negative number, we consider number of houses, \(x = 35\)

Therefore, house number \(35\) is such that the sum of the numbers of houses preceding the house numbered \(35\) is equal to the sum of the numbers of the houses following it.