Arithmetic Progressions class 10 exercise 5.4 NCERT Solutions PDF
Neha Tanna24 Nov, 2025
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NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.4: Arithmetic Progressions class 10 exercise 5.4 provide a structured and exam-oriented approach to mastering Arithmetic Progressions, an important chapter in the CBSE Class 10 syllabus.
This exercise helps students strengthen core AP concepts such as identifying terms, finding the nth term, and calculating the sum of n terms.
With clear explanations and precise steps, these solutions enhance conceptual clarity, improve accuracy, and support effective preparation for the Class 10 Maths Board Exam.
Class 10 Maths Chapter 5 Exercise 5.4 Arithmetic Progressions Solutions
These solutions are designed to help students understand every question in the Arithmetic Progressions Class 10 Exercise 5.4 with clarity and accuracy.1. Which term of the AP: 121, 117, 113,..., is its first negative term? [Hint: Find n for a n < 0]
Solution:
Given the AP series is 121, 117, 113,..., thus, the first term, a = 121. The common difference, d = 117-121= -4By the nth term formula, a n = a +( n −1) d. Therefore, a n = 121+(n−1)(-4) = 121-4n+4 =125-4n To find the first negative term of the series, a n < 0.
Therefore, 125-4n < 0 125 < 4n n>125/4 n>31.25
Therefore, the first negative term of the series is the 32 nd 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:
From the given statements, we can write a 3 + a 7 = 6 …………………………….(i)By the nth term formula, a n = a +( n −1) d Third term, a 3 = a+(3 -1)d a 3 = a + 2d………………………………(iii)
And Seventh term, a7= a+(7-1)d a 7 = a + 6d ………………………………..(iv)
From equations (iii) and (iv), putting in equation(i), we get a+2d +a+6d = 6 2a+8d = 6 a+4d=3 or a = 3–4d …………………………………(v)
Putting the value of a from equation (v), we get (3–4d +2d)×(3–4d+6d) = 8 (3 –2d)×(3+2d) = 8 3 2 – 2d 2 = 8 9 – 4d 2 = 8 4d 2 = 1 d = 1/2 or -1/2
Now, by putting both the values of d, we get a = 3 – 4d = 3 – 4(1/2) = 3 – 2 = 1, when d = 1/2 a = 3 – 4d = 3 – 4(-1/2) = 3+2 = 5, when d = -1/2
We know that the sum of the nth term of AP is S n = n /2 [2 a +( n – 1) d ]. So, when a = 1 and d=1/2
Then, the sum of the first 16 terms is S 16 = 16 /2 [2 +(16-1)1/2] = 8(2+15/2) = 76
And when a = 5 and d= -1/2 Then, the sum of the first 16 terms is S 16 = 16 /2 [2(5)+(16-1)(-1/2)] = 8(5/2)=20
3. 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
apart, what is the length of the wood required for the rungs? [Hint: Number of rungs = -250/25 ]. 
Solution:
Given, The distance between the rungs of the ladder is 25cm. Distance between the top rung and bottom rung of the ladder is =
= 5/2 ×100cm = 250cm Therefore, total number of rungs = 250/25 + 1 = 11.
As we can see from the figure, the ladder has rungs in decreasing order from top to bottom.
Thus, we can conclude that the rungs are decreasing in the order of AP. And the length of the wood required for the rungs will be equal to the sum of the terms of the AP series formed.
So, The first term, a = 45 The last term, l = 25 Number of terms, n = 11
Now, as we know, the sum of nth terms is equal to S n = n/2(a+ l ) S n = 11/2(45+25) = 11/2(70) = 385 cm. Hence, the length of the wood required for the rungs is 385cm.
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. [Hint :Sx – 1 = S49 – Sx ]
Solution:
Given, Row houses are numbers from 1,2,3,4,5…….49.Thus, we can see the houses numbered in a row are in the form of AP. So, The first term, a = 1 The common difference, d=1
Let us say the number of the houses can be represented as Sum of nth term of AP = n/2[2a+(n-1)d]
Sum of number of houses beyond x house = S x-1 = (x-1)/2[2(1)+(x-1-1)1] = (x-1)/2 [2+x-2] = x(x-1)/2 ………………………………………(i)
By the given condition, we can write S 49 – S x = {49/2[2(1)+(49-1)1]}–{x/2[2(1)+(x-1)1]} = 25(49) – x(x + 1)/2 ………………………………….(ii)
As per the given condition, eq.(i) and eq(ii) are equal to each other. Therefore, x(x-1)/2 = 25(49) – x(x+1)/2 x = ±35.
As we know, the number of houses cannot be a negative number.
Hence, the value of x is 35.
5. 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 1 4 m and a tread of 1 2 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 ×1/2 ×50 m 3 .]
Solution:
As we can see from the given figure, the first step is ½ m wide, 2 nd step is 1m wide, and 3 rd step is 3/2m wide.Thus, we can understand that the width of the step by ½ m each time when the height is ¼ m. And also, given the length of the steps is 50m, the time.
So, the width of steps forms a series AP in such a way that ½ , 1, 3/2, 2, ……..
The volume of steps = Volume of Cuboid = Length × Breadth Height.
Now, The volume of concrete required to build the first step = ¼ ×1/2 ×50 = 25/4
The volume of concrete required to build the second step =¼ ×1×50 = 25/2
The volume of concrete required to build the second step = ¼ ×3/2 ×50 = 75/4
Now, we can see the volumes of concrete required to build the steps are in the AP series. 25/4 , 25/2 , 75/4 …..
Thus, applying the AP series concept, The first term, a = 25/4 The common difference, d = 25/2 – 25/4 = 25/4
As we know, the sum of n terms is S n = n/2[2a+(n-1)d] = 15/2(2×(25/4 )+(15/2 -1)25/4) Upon solving, we get S n = 15/2 (100) S n =750
Hence, the total volume of concrete required to build the terrace is 750 m 3 .
NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.4 PDF
The Class 10 Arithmetic Progressions Exercise 5.4 PDF provides clear, step-by-step solutions to all questions based on the CBSE syllabus. It covers important concepts like the nth term and sum of an AP, helping students revise efficiently and strengthen problem-solving skills. This PDF is ideal for quick practice and reliable exam preparation.
NCERT solutions Class 10 Maths PDF
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