NCERT Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.4

NCERT Solutions Class 10 Maths Chapter 5 Exercise 5.4 will delve into the solutions of the optional exercise of Chapter 5 on Arithmetic Progressions (AP). The exercise is focused on further solidifying the concepts of AP, which were studied in previous exercises. This exercise is all about real-life scenarios of APs and calculating their sums or terms. Here, we have explained the details of  Exercise 5.4 to help students tackle complex problems and applications, making these mathematical ideas more relevant and accessible. So, let’s dive deeper into this exercise. 

1.0Download NCERT Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.4 : Free PDF

2.0Introduction to Arithmetic Progressions (AP)

In mathematics, an Arithmetic Progression (AP) is a sequence of numbers such that the difference between consecutive terms is always the same. This is known as the common difference (d). For instance, in the series 3, 7, 11, 15,…, the common difference is 4 because every term is obtained by adding 4 to the preceding term. Note that the common difference is what differentiates an AP from other types of sequences. This is because the common difference of an Arithmetic Progression remains constant throughout the sequence. 

3.0Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.4: Overview

The nth Term of an Arithmetic Progression (AP):

The nth term of an arithmetic progression is the term occurring at the n-th position within the sequence. It is any term in the sequence, and its value can be determined with the help of the first term and the difference between consecutive terms. The nth term formula allows you to find the value of any term in the sequence from its position. The formula to find the nth term of an AP is: 

$a_n=a+(n-1)d$

Here, 

• a_n = nth term of an AP. 
• a = first term of an AP
• n = number of terms in an AP.
• d = Common Difference calculated by subtracting subsequent terms from the previous term of the AP. 

The sum of the First n Terms of an AP

The sum of the first n terms of an arithmetic progression (AP) is the total value that is achieved by adding the first n terms of the sequence. The sum is used to determine the total value of a given set of terms in the AP without having to add them separately. The sum is based on the first term, the common difference, and the number of terms taken. The formula to find the sum of n terms of AP can be expressed as: 

$S_n=\frac{n}{2}\left[2a+(n-1)d\right]$

Here, 

• S_n​ = sum of the first n terms
• a = first term of the AP
• d = common difference
• n = number of terms

If the first and the last term(l or a_n) of an AP is given, then the formula to find the sum of this AP can be written as: 

$S_n=\frac{n}{2}(a+a_n)$

Remark: If given that an AP contains, a, b, and c as its terms, then the relation between these three terms can be expressed as: 

$b=\frac{a+c}{2}$

The relation mentioned here is also known as the arithmetic mean of a and c with respect to b. 

Also Read: NCERT Solutions for Maths Chapter 10

4.0NCERT Class 10 Maths Chapter 5 Arithmetic Progression Exercise 5.4 : Detailed Solutions

1. Which term of the AP : 121, 117, 113, ..., is its first negative term?

Solution:

121, 117, 113, … an < 0

=> a + (n - 1)d < 0

=> 121 + (n - 1)(-4) < 0

=> 121 - 4n + 4 < 0

=> 125 < 4n

=> n > 125/4 ≈ 31.25

The first negative term should be the 32nd term.

2. The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of the first sixteen terms of the AP.

Solution:

Given, a3 + a7 = 6 and (a3)(a7) = 8

a + 2d + a + 6d = 6

2a + 8d = 6

a + 4d = 3

=> a = 3 - 4d

Also, (a + 2d)(a + 6d) = 8

=> [3 - 4d + 2d][3 - 4d + 6d] = 8

=> [3 - 2d][3 + 2d] = 8

=> 9 - 4d² = 8

=> 4d² = 1

=> d = ± 1/2

When d = 1/2, a = 3 - 4(1/2) = 1

When d = -1/2, a = 3 - 4(-1/2) = 5

When, a = 1, d = 1/2

=> S16 = (16/2)[2 × 1 + (16 - 1) × (1/2)]

= 8[2 + 15/2] = 8 × 19/2 = 76

When a = 5, d = -1/2

=> S16 = (16/2)[2 × 5 + (16 - 1)(-1/2)]

= 8[10 - 15/2] = 8 × 5/2 = 20

∴ S16 = 20 or 76

3. A ladder has rungs 25 cm apart. 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 1/2 m apart, what is the length of the wood required for the rungs?

Solution:

Number of rungs = (250/25) + 1 = 11

Largest rung = 45 cm

Smallest one = 25 cm

a = 45 cm

Length of last rung = 25 cm

Now, length of wood required to form 11 rungs

= (n/2)(a + l)

= (11/2)(45 + 25) cm

= (11/2 × 70) cm = 385 cm

4. 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:

Here, the AP form is 1, 2, 3, ..., 49.

Here, sum of the numbers of houses preceding the house numbered x = sum of first (x - 1) numbered houses = Sx-1

Sum of numbers of houses following house numbered x = sum of 49 numbered houses - sum of x numbered houses = S49 - Sx

According to the question,

Sx-1 = S49 - Sx

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

=> [(x - 1)/2]x = (49/2) × 50 - (x/2)(x + 1)

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

=> (x/2) × 2x = x² = 1225

=> x = √1225 = 35

5. A small terrace at a football ground comprises 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of 1/4 m and a tread of 1/2 m. Calculate the total volume of concrete required to build the terrace.

Solution:

Given number of steps, n = 15

Tread breadth = 1/2 m

Length = 50 m

Now, volume of first step = (1/4) × (1/2) × 50

Volume of second step = (1/4 + 1/4) × (1/2) × 50 = (2/4) × (1/2) × 50

Volume of third step = (3/4) × (1/2) × 50

Hence the series formed as:

[(1/4) × (1/2) × 50] + [(2/4) × (1/2) × 50] + [(3/4) × (1/2) × 50] + ... (15 terms)

Or (1/2) × 50[(1/4) + (2/4) + (3/4) + ...]

Or (1/2) × 50 [(15/2)[2 × (1/4) + (15 - 1) × (1/4)]]

=> 25 × (15/2)[(2/4) + (14/4)] = 25 × (15/2) × (16/4)

=> 25 × 15 × 2 = 750 m³

5.0Benefits of NCERT Solutions Class 10 Maths Chapter 5 Exercise 5.4

• Provides step-by-step explanations, making it easier to understand AP concepts thus increasing conceptual clarity
• Helps students prepare effectively for board exams
• Strengthens understanding of key formulas for the nth term and sum of AP terms.
• Well-structured solutions allow quick and efficient revision before exams.
• Step-by-step solutions make complex problems easy to solve, increasing student confidence.