---
title: "NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.4"
url: https://www.swavid.com/maths/class/10/chapter/arithmetic-progressions/ncert-solutions/exercise-5-4
dateModified: 2026-10-07T15:51:43+00:00
---

# NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.4

Chapter 5: Arithmetic Progressions. Every question from Exercise 5.4, with full working and the final answer.

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## EXERCISE 5.4 (Optional)*

### Question 1

*3 marks · Short answer*

Which term of the AP : $121, 117, 113, \dots$, is its first negative term?

**Solution**

1. Given the AP: $121, 117, 113, \dots$, the first term $a = 121$ and common difference $d = 117 - 121 = -4$.
2. We need to find the first negative term, so we set $a_n < 0$, which gives $a + (n - 1)d < 0$.
3. Substituting the values, $121 + (n - 1)(-4) < 0$, which simplifies to $121 - 4n + 4 < 0$ or $125 < 4n$.
4. This means $4n > 125$, so $n > \frac{125}{4} = 31.25$.
5. Since $n$ must be a positive integer, the smallest integer greater than $31.25$ is $32$.
6. Therefore, the 32nd term is the first negative term.

**Answer:** 32nd term

> Common mistake: Taking $n$ as 31 instead of rounding up to the next integer 32.

### Question 2

*3 marks · Short answer*

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.

**Solution**

1. Let the first term be $a$ and the common difference be $d$.
2. Given that the sum of the third and seventh terms is $6$, we have $(a + 2d) + (a + 6d) = 6$, which simplifies to $2a + 8d = 6$ or $a + 4d = 3$.
3. Given that their product is $8$, we have $(a + 2d)(a + 6d) = 8$.
4. From the first equation, $a = 3 - 4d$. Substituting this into the product equation gives $(3 - 4d + 2d)(3 - 4d + 6d) = 8$, or $(3 - 2d)(3 + 2d) = 8$.
5. This yields $9 - 4d^2 = 8$, so $4d^2 = 1$, giving $d = \frac{1}{2}$ or $d = -\frac{1}{2}$.
6. Case 1: When $d = \frac{1}{2}$, $a = 3 - 4(\frac{1}{2}) = 1$. The sum of the first 16 terms is $S_{16} = \frac{16}{2}[2(1) + (16 - 1)(\frac{1}{2})] = 8[2 + \frac{15}{2}] = 8(\frac{19}{2}) = 76$.
7. Case 2: When $d = -\frac{1}{2}$, $a = 3 - 4(-\frac{1}{2}) = 5$. The sum of the first 16 terms is $S_{16} = \frac{16}{2}[2(5) + (16 - 1)(-\frac{1}{2})] = 8[10 - \frac{15}{2}] = 8(\frac{5}{2}) = 20$.

**Answer:** 76 or 20

> Common mistake: Dropping one of the possible values for the common difference $d$.

### Question 3

*3 marks · Short answer*

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 the bottom rungs are $2\frac{1}{2}$ m apart, what is the length of the wood required for the rungs?

**Solution**

1. Given: distance between rungs is $25\text{ cm}$, bottom rung length is $45\text{ cm}$, top rung length is $25\text{ cm}$, and total height is $2\frac{1}{2}\,\text{m} = 250\text{ cm}$.
2. The number of rungs is given by $\text{Number of rungs} = \frac{250}{25} + 1 = 11$.
3. The lengths of the rungs form an AP with first term $a = 45$, last term $l = 25$, and number of terms $n = 11$.
4. The total length of the wood required is the sum of 11 terms of this AP: $S_{11} = \frac{n}{2}(a + l) = \frac{11}{2}(45 + 25) = \frac{11}{2} \times 70 = 385\text{ cm}$.

**Answer:** $385\text{ cm}$

> Common mistake: Forgetting to add 1 when calculating the number of rungs, leading to $10$ rungs instead of $11$.

### Question 4

*3 marks · Short answer*

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 $x$.

**Solution**

1. The house numbers are $1, 2, 3, \dots, 49$, which form an AP with $a = 1$ and $d = 1$.
2. Let the house number be $x$. The sum of the house numbers preceding $x$ is $S_{x-1} = \frac{x-1}{2}[2(1) + (x - 1 - 1)(1)] = \frac{(x-1)x}{2}$.
3. The sum of the house numbers following $x$ is equal to the total sum of all 49 houses minus the sum of the first $x$ houses, which is $S_{49} - S_x$.
4. Given that $S_{x-1} = S_{49} - S_x$, we can rewrite this as $S_{x-1} + S_x = S_{49}$.
5. Using the formula for the sum of $n$ terms, $\frac{(x-1)x}{2} + \frac{x(x+1)}{2} = \frac{49(50)}{2}$.
6. Simplifying gives $x^2 - x + x^2 + x = 49 \times 50$, so $2x^2 = 2450$, which means $x^2 = 1225$.
7. Taking the positive square root, $x = 35$.

**Answer:** $35$

> Common mistake: Setting $S_{x-1} = S_x$ instead of equating the sum before $x$ to the sum after $x$.

### Question 5

*3 marks · Short answer*

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 $\frac{1}{4}$ m and a tread of $\frac{1}{2}$ m. (see Fig. 5.8). Calculate the total volume of concrete required to build the terrace.

**Solution**

1. The volume of concrete for the first step is given by $\frac{1}{4} \times \frac{1}{2} \times 50 = \frac{25}{4}\text{ m}^3$.
2. The volume of concrete for the second step is $\frac{2}{4} \times \frac{1}{2} \times 50 = \frac{50}{4}\text{ m}^3$, and for the third step is $\frac{3}{4} \times \frac{1}{2} \times 50 = \frac{75}{4}\text{ m}^3$.
3. The volumes of concrete required for the 15 steps form an AP with first term $a = \frac{25}{4}$ and common difference $d = \frac{50}{4} - \frac{25}{4} = \frac{25}{4}$.
4. The total volume of concrete is the sum of the first 15 terms of this AP, $S_{15} = \frac{15}{2}[2a + (15 - 1)d]$.
5. Substituting the values, $S_{15} = \frac{15}{2}[2(\frac{25}{4}) + 14(\frac{25}{4})] = \frac{15}{2} \times \frac{25}{4} \times (2 + 14) = \frac{15}{2} \times \frac{25}{4} \times 16$.
6. Simplifying gives $S_{15} = 15 \times 25 \times 2 = 750\text{ m}^3$.

**Answer:** $750\text{ m}^3$

> Common mistake: Arithmetic errors when multiplying fractions in the sum formula.

## Related pages

- [All Chapter 5 NCERT solutions](https://www.swavid.com/maths/class/10/chapter/arithmetic-progressions/ncert-solutions)
- [Exercise 5.1](https://www.swavid.com/maths/class/10/chapter/arithmetic-progressions/ncert-solutions/exercise-5-1)
- [Exercise 5.2](https://www.swavid.com/maths/class/10/chapter/arithmetic-progressions/ncert-solutions/exercise-5-2)
- [Exercise 5.3](https://www.swavid.com/maths/class/10/chapter/arithmetic-progressions/ncert-solutions/exercise-5-3)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student