In the verge of coronavirus pandemic, we are providing FREE access to our entire Online Curriculum to ensure Learning Doesn't STOP!

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

Go back to  'Ex.5.4'

## 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: $${S_x}_{ - 1} = {S_{49}} - {S_x}$$]

Video Solution
Arithmetic Progressions
Ex 5.4 | Question 4

## 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.

Learn from the best math teachers and top your exams

• Live one on one classroom and doubt clearing
• Practice worksheets in and after class for conceptual clarity
• Personalized curriculum to keep up with school