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

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## 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} &= \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]\\ &= \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.

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