# 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 Sₓ ₋ ₁ = S₄₉ - Sₓ ]

**Solution:**

An arithmetic progression is a sequence that has a common difference between any two of its consecutive numbers.

The sum of the first n terms of an AP is given by Sₙ = n/2 [2a + (n - 1) d], where a is the first term, d is a common difference and n is the number of terms.

The number of houses is 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^{th} house can be expressed as below:

We know that sum of n terms in an A.P. is given by the formula Sₙ = n/2 [2a + (n - 1) d]

Sum of number of houses preceding x^{th} house = Sₓ ₋ ₁

Sₓ ₋ ₁_{ }= (x - 1) / 2 [2a + ((x - 1) - 1)d]

= (x - 1) / 2 [2 × 1 + ( x - 2) × 1]

= (x - 1) / 2 [2 + x - 2]

= [x (x - 1)] / 2 ---------- (1)

By the given information we know that, sum of number of houses following x^{th} house = S₄₉ - Sₓ

S₄₉ - Sₓ = 49 / 2 [2 × 1 + (49 - 1) × 1] - x / 2 [2 × 1 + (x - 1) × 1]

= 49 / 2 [2 + 48] - x / 2 (2 + x - 1)

= (49 / 2) × 50 - (x / 2) (x + 1)

= 1225 - [x (x + 1)] / 2 ---------- (2)

It is given that these sums are equal, that is equation (1) = equation(2)

x (x - 1) / 2 = 1225 - x (x + 1) / 2

x² / 2 - x / 2 = 1225 - x² / 2 - x / 2

On solving further we get,

x^{2} = 1225

x = ± 35

As the number of houses cannot be a negative number, we consider the number of houses as x = 35

**☛ Check: **NCERT Solutions Class 10 Maths Chapter 5

**Video Solution:**

## 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.

Class 10 Maths NCERT Solutions Chapter 5 Exercise 5.4 Question 4

**Summary:**

The houses of a row are numbered consecutively from 1 to 49. We have shown 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. The value of x is 35.

**☛ Related Questions:**

- Which term of the A.P : 121, 117, 113, ..., is its first negative term?[Hint : Find n for an less than o]
- The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.
- A ladder has rungs 25 cm apart. (see Fig. 5.7). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and bottom rungs are 2 ½ m apart, what is the length of the wood required for the rungs? [Hint: number of rungs = 250 / 25 + 1]
- A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of ¼ m and a tread of ½ m . (see Fig. 5.8). Calculate the total volume of concrete required to build the terrace. [Hint : Volume of concrete required to build the first step =¼ × ½ × 50 m³.].

visual curriculum