Class 10 Maths Chapter 5, Arithmetic Progressions, introduces sequences with a constant common difference d, derives the nth term formula a_n = a + (n – 1)d and the sum of the first n terms S_n = n/2[2a + (n – 1)d] = n/2(a + l), and applies them to real-world patterns.
What Class 10 Mathematics (Ganit — Hindi Medium) Chapter 5 Covers
5.1 & 5.2 Arithmetic Progressions & nth Term of an AP defines an Arithmetic Progression (AP) with first term a and common difference d = ak+1 – a_k, writes the general form a, a+d, a+2d, \dots, and derives the general term a_n = a + (n – 1)d (including finding the kth term from the end of a finite AP).
5.3 Sum of First n Terms of an AP recounts Gauss’s reversal trick for 1 + 2 + \dots + 100 = 5050 to derive S_n = n/2[2a + (n – 1)d] = n/2(a + a_n), notes that a_n = S_n – Sn-1, and shows that the sum of the first n positive integers is n(n+1)/2.
Exercises in Chapter 5
- Exercise 5.1 (4 Questions) checks which lists of numbers form an AP and finds the first term a, common difference d and next three terms.
- Exercise 5.2 (20 Questions) applies a_n = a + (n – 1)d to find missing terms, count three-digit numbers divisible by 7, and solve salary/savings problems.
- Exercise 5.3 (20 Questions) applies S_n to sums of series, penalty-for-delay construction contracts, cash prizes, tree-planting drives, spiral lengths, stacked logs and the potato race.
Read Class 10 Mathematics (Ganit — Hindi Medium) Chapter 5 Online
More Chapters
- Previous: Chapter 4: द्विघात समीकरण (Quadratic Equations)
- Next: Chapter 6: त्रिभुज (Triangles)
- English medium: Class 10 Maths Chapter 5 PDF in English
- All Mathematics (Ganit — Hindi Medium) chapters and the other Class 10 books: Class 10 NCERT books 2026-27
- All Class 1 to 12 NCERT Textbooks: NCERT Books PDF (Class 1–12)
- Every book chapter by chapter: NCERT chapter wise PDF
FAQ
What are the formulas for the nth term and sum of n terms of an AP?
With first term a, common difference d and last term l = a_n: the nth term is a_n = a + (n – 1)d, and the sum of the first n terms is S_n = n/2[2a + (n – 1)d] = n/2(a + l).
How do you find the nth term a_n if the sum S_n of the first n terms is given?
For $n = 1$, a1 = S1; for any n ≥ 2, the nth term is a_n = S_n – Sn-1.