NCERT Solutions for Class 10 Maths Chapter 5: Arithmetic Progressions
Students searching for accurate and easy-to-follow NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions will find everything they need at Myclass24. Arithmetic Progressions, commonly known as AP, is one of the most rewarding chapters in Class 10 Maths because the formulas are few but the application range is wide. Once students understand the structure of an AP — the first term (a), the common difference (d), and the nth term (aₙ) — solving even complex problems becomes straightforward. Myclass24's solutions cover all four exercises of Chapter 5 with detailed workings and exam-focused explanations.
Whether it's finding the sum of the first 20 terms or determining whether a given list of numbers forms an AP, the solutions are laid out step-by-step for maximum clarity. Chapter 5 also features some of the most interesting word problems in the entire Class 10 Maths book, including problems on savings, stacking, seating arrangements, and even sports applications. These make the chapter both practically relevant and exam-scoring.
Download PDF – NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions
The PDF of NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions is freely available on Myclass24. It covers Exercise 5.1, Exercise 5.2, Exercise 5.3, and Exercise 5.4 (optional). The PDF is formatted to be mobile-friendly, with clear formulas and step-wise solutions that load well on smartphones and tablets.
Chapter 5 Arithmetic Progressions – Core Formulas, Facts & Exercise Breakdown
NCERT Class 10 Arithmetic Progression is a list of numbers where the difference between any two consecutive terms is constant. This constant difference is called the common difference (d). If the first term is a and the common difference is d, then the nth term of the AP is given by: aₙ = a + (n–1)d. This formula allows finding any term of an AP directly without listing all previous terms. The sum of the first n terms is given by: Sₙ = n/2 [2a + (n–1)d] or alternatively Sₙ = n/2 (a + l), where l is the last term. A critical relationship is aₙ = Sₙ – Sₙ₋₁, which means the nth term equals the difference of the sum of n terms and sum of (n-1) terms. This identity appears in many tricky CBSE problems.
| Exercise | Topics Covered | No. of Questions | Difficulty Level |
| Exercise 5.1 | Introduction to AP, identifying AP from list | 4 | Easy |
| Exercise 5.2 | nth Term Formula (aₙ = a + (n–1)d) | 20 | Easy–Medium |
| Exercise 5.3 | Sum of n Terms (Sₙ formula) | 20 | Medium–Hard |
| Exercise 5.4 | Optional – Higher-Order AP Problems | 5 | Hard |
| Formula | Expression | Used For |
| nth Term | aₙ = a + (n–1)d | Finding any specific term |
| Sum of n Terms (Form 1) | Sₙ = n/2 [2a + (n–1)d] | When first term & d are known |
| Sum of n Terms (Form 2) | Sₙ = n/2 (a + l) | When first & last term are known |
| Common Difference | d = aₙ – aₙ₋₁ | Verifying AP; finding d |
| Number of Terms | n = (l – a)/d + 1 | Finding total terms in a finite AP |
| Middle Term (odd n) | a(n+1)/2 | Middle term of finite AP |
One fact that surprises many students: if three numbers a, b, c are in AP, then 2b = a + c. This simple property is used in problems where you are asked to find three or four numbers in AP given their sum and product. Another frequently tested concept is finding the sum of the first n natural numbers, first n even numbers, and first n odd numbers — all derivable from the Sₙ formula. Exercise 5.3 is the most important exercise in this chapter and is filled with word problems involving money saved over days, logs stacked in rows, seats arranged in a theatre, and similar real-life APs.
| Question Type | Typical Marks | Frequency in Board Exams |
| Find nth term of an AP | 1–2 Marks | Very High |
| Find number of terms in an AP | 2 Marks | High |
| Sum of first n terms | 3 Marks | Very High |
| Word problem using AP sum | 4 Marks | High |
| Find a and d given two conditions | 3–4 Marks | High |
| Check if list of numbers is AP | 1 Mark | Medium (MCQ) |