NCERT Solutions for Chapter 5 of Class 10 Maths, Arithmetic Progressions, introduces students to the interesting world of sequences and patterns that can be found in everyday objects like pine cone spirals and flower petals. This chapter explores how each term in a sequence can be derived by adding a fixed number, known as the common difference, to the previous term. Understanding these concepts is crucial for solving real-life problems.
The solutions provide a comprehensive overview of key terminologies, including 'term' and 'common difference', along with detailed explanations of how to find the 'n-th' term and the sum of 'n' consecutive terms. NCERT Class 10 Maths - Chapter 5 provides students with well-illustrated examples and formulas to help them better understand arithmetic progressions. This will help them cope with exercises and perform better on exams.
The NCERT Solutions for Class 10 Maths Chapter 5 is available in the downloadable PDF format below:
Below we have provided the list of topics that we will cover in the NCERT Solution Class 10 Chapter 5:
The NCERT solutions offer clear answers to all exercises, helping students solve each problem confidently. By learning about concepts like common differences, the n-th term, and the sum of terms, students can prepare well for exams and gain a better understanding of how arithmetic progressions are used in math and real-life situations. Check the number of questions included in each exercise.
There are a few reasons that will help you understand why you need to choose the NCERT Solution for Class 10 Math:
Chapter 5, Arithmetic Progression (AP), deals with sequences where the difference between consecutive terms is constant, known as the common difference (d). It covers important concepts like finding the nth term of an AP and the sum of the first n terms. Students also learn to apply these concepts in real-life problems, such as calculating distances or total amounts. Mastering AP helps in solving various mathematical problems and understanding sequences in algebra.
Definition: An arithmetic progression (AP) is a sequence of numbers in which the difference between consecutive terms is constant. This difference is called the common difference.
General Form of an AP: The nth term of an AP is given by: where:
Examples of AP:
Sum of First n Terms of an AP
Formula for the sum of the first n terms of an arithmetic progression:
or alternatively:
where l is the last term of the AP.
Application: This formula is used to calculate the sum of a series of numbers in an arithmetic progression.
1. In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?
(i) The taxi fare after each km when the fare is ₹ 15 for the first km and ₹ 8 for each additional km.
(ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.
(iii) The cost of digging a well after every metre of digging, when it costs ₹ 150 for the first metre and rises by ₹ 50 for each subsequent metre.
(iv) The amount of money in the account every year, when ₹ 10,000 is deposited at compound interest at 8% per annum.
Solution
(i) Let tₙ denote the taxi fare (in ₹) for the first n km.
Now, t₁ = 15,
t₂ = 15 + 8 = 23,
t₃ = 23 + 8 = 31,
t₄ = 31 + 8 = 39, ...
List of fares after 1 km, 2 km, 3 km, 4 km, ... respectively is 15, 23, 31, 39, .... (in ₹).
Here, t₂ -t₁ = t₃ - t₂ = t₄ - t₃ = ... = 8. Thus, the list forms an AP.
(ii) Let t₁ = x units; t₂ = x - (1/4)x = (3/4)x units;
t₃ = (3/4)x - (1/4)(3/4)x = (9/16)x units;
t₄ = (9/16)x - (1/4)(9/16)x = (27/64)x units; ...
The list of numbers is x, (3/4)x, (9/16)x, (27/64)x, ....
It is not an AP because t₂ - t₁ ≠t₃ - t₂.
(iii) Cost of digging for first metre = ₹ 150
Cost of digging for first 2 metres = 150 + 50 = ₹ 200
Cost of digging for first 3 metres = 200 + 50 = ₹ 250
Cost of digging for first 4 metres = 250 + 50 = ₹ 300
Clearly, 150, 200, 250, 300.... forms an AP.
Here, t₂ - t₁ = t₃ - t₂ = t₄ - t₃ = ... . Thus, the list forms an AP.
(iv) We know that if ₹ P is deposited at r% compound interest per annum for n years, our money will be P(1 + r/100) after n years.
Therefore, after every year, our money will be
10000(1 + 8/100), 10000(1 + 8/100)²,
10000(1 + 8/100)³, 10000(1 + 8/100)⁴, ...
Clearly, adjacent terms of this series do not have the same difference between them. Therefore, this is not an AP.
2. Fill in the blanks in the following table, given that a is the first term, d the common difference and , the nth term of the AP.
Solution:
(i) a = 7, d = 3, n = 8
a₈ = a + 7d = 7 + 7 × 3 = 28.
Hence, a₈ = 28.
(ii) a = -18, n = 10, aₙ = 0, d = ?
aₙ = a + (n - 1)d
0 = -18 + (10 - 1)d
18 = 9d => d = 18/9 = 2
Hence, d = 2.
(iii) d = -3, n = 18, aₙ = -5
aₙ = a + (n - 1)d
-5 = a + (18 - 1)(-3)
-5 = a + (17)(-3)
-5 = a - 51
a = 51 - 5 = 46
Hence, a = 46.
(iv) a = -18.9, d = 2.5, aₙ = 3.6
=> a + (n - 1)d = 3.6
=> -18.9 + (n - 1) × (2.5) = 3.6
=> (n - 1) × (2.5) = 3.6 + 18.9 = 22.5
=> n - 1 = 22.5 / 2.5 = 225 / 25 = 9
=> n = 10
(v) a = 3.5, d = 0, n = 105
Then a₁₀₅ = a + 104d = 3.5 + 0 = 3.5
3. For the following APs, write the first term and the common difference
(i) 3, 1, -1, -3, ...
(ii) -5, -1, 3, 7, ...
(iii) 1/3, 5/3, 9/3, 13/3, ...
(iv) 0.6, 1.7, 2.8, 3.9, ...
Solution
(i) a = 3, d = t₂ - t₁ = 1 - 3 = -2, i.e., d = -2
(ii) a = -5, d = t₂ - t₁ = -1 - (-5) = 4
(iii) a = 1/3, d = t₂ - t₁ = 5/3 - 1/3 = 4/3
(iv) a = 0.6, d = t₂ - t₁ = 1.7 - 0.6 = 1.1
4. Ramkali saved ₹ 5 in the first week of a year and then increased her weekly savings by ₹ 1.75. If in the nth week, her weekly savings become ₹ 20.75, find n.
Solution
= ₹ 5 (savings in the 1st week)
= ₹ 5 + ₹ 1.75 = ₹ 6.75 (savings in the 2nd week)
= ₹ 6.75 + ₹ 1.75 = ₹ 8.50 (savings in the 3rd week)
= ₹ 20.75
=> a + (n - 1)d = 20.75
=> 5 + (n - 1) × 1.75 = 20.75
=> (n - 1) × 1.75 = 15.75
=> n - 1 = 15.75 / 1.75 = 1575 / 175 = 9
=> n = 10
Hence, in the 10th week, Ramkali's savings will be ₹ 20.75.
5. Subba Rao started work in 1995 at an annual salary of ₹ 5000 and received an increment of ₹ 200 each year. In which year did his income reach ₹ 7000?
Solution
It can be observed that the incomes that Subba Rao obtained in various years are in AP as every year, his salary is increased by ₹ 200.
Therefore, the salaries of each year after 1995 are
5000, 5200, 5400, ...
Here, a = 5000
d = 200
Let after nth year, his salary be ₹ 7000.
Therefore, an = a + (n - 1)d
7000 = 5000 + (n - 1)200
200(n - 1) = 2000
(n - 1) = 10
n = 11
Therefore, in the 11th year, his salary will be ₹ 7000.
Arithmetic Progression (AP) offers several benefits:
Overall, AP is easy to work with and has practical uses in finance, education, and real-world scenarios.
(Session 2025 - 26)