Arithmetic Progressions – Class 10 Maths Chapter 5 Notes & NCERT PDF

Class 10MathematicsChapter 5NCERT book: Mathematics - Class 10

Notes and a simple summary of Chapter 5 of the NCERT Class 10 Mathematics book, with the official chapter PDF, flashcards, an MCQ quiz and an AI tutor for your doubts.

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Chapter 5: Arithmetic Progressions

Introduction

The chapter opens with patterns from daily life. Reena's salary starts at ₹8000 with an annual increment of ₹500, so her yearly salary is 8000, 8500, 9000, …. The rungs of a ladder decrease uniformly by 2 cm from a 45 cm bottom rung: 45, 43, 41, …, 31. In each list, every term is obtained by adding a fixed number to the previous term. Such a list is called an Arithmetic Progression (AP).

Key Concepts & Formulas

What is an AP?

An arithmetic progression is a list of numbers in which each term is obtained by adding a fixed number to the preceding term, except the first term. This fixed number is called the common difference (d).

General form: a, a + d, a + 2d, a + 3d, …

d can be positive (increasing AP), negative (decreasing AP) or zero (all terms equal).

Test for an AP: a list a₁, a₂, a₃, … is an AP if aₖ₊₁ − aₖ is the same for every k.

A finite AP has a last term; an infinite AP does not.

nth term (general term)

aₙ = a + (n − 1)d

The last term of a finite AP is often written l, so l = a + (n − 1)d.

To find the nth term from the end, reverse the AP (first term l, common difference −d), or first find the total number of terms.

Sum of the first n terms

Sₙ = n/2 [2a + (n − 1)d]

Sₙ = n/2 (a + l) when the last term l is known

Sum of the first n positive integers: Sₙ = n(n + 1)/2 (so 1 + 2 + … + 1000 = 500500).

The book tells the story of Gauss, who as a child added 1 to 100 by pairing 1 + 100, 2 + 99, … — each pair gives 101, and there are 50 pairs: 5050.

aₙ = Sₙ − Sₙ₋₁ (the nth term from the sums).

Methods Step by Step

Is it an AP? Compute a₂ − a₁, a₃ − a₂, a₄ − a₃. Same value → AP, and that value is d.

Which term is x? Put aₙ = x in a + (n − 1)d = x and solve for n. If n is not a positive integer, x is not a term.

Find the AP from two terms: write aₚ = a + (p − 1)d and a_q = a + (q − 1)d, then solve the pair of linear equations.

How many terms give sum S? Put Sₙ = S, get a quadratic in n, and keep only positive integer roots.

Worked Examples (from the chapter)

10th term of 2, 7, 12, … — a = 2, d = 5: a₁₀ = 2 + 9 × 5 = 47.

Which term of 21, 18, 15, … is −81? −81 = 21 + (n − 1)(−3) gives n = 35. Setting aₙ = 0 gives n = 8, so the 8th term is 0.

Is 301 a term of 5, 11, 17, 23, …? 301 = 5 + (n − 1)6 gives n = 302/6 = 151/3, not a whole number, so 301 is not a term.

Two-digit numbers divisible by 3: 12, 15, …, 99 → 99 = 12 + (n − 1)3 gives n = 30.

11th term from the end of 10, 7, 4, …, −62: there are 25 terms, so it is the 15th term: 10 + 14(−3) = −32.

Simple interest: ₹1000 at 8% gives yearly interest 80, 160, 240, … — an AP with a = d = 80, so a₃₀ = ₹2400.

Sum of first 22 terms of 8, 3, −2, … = 11[16 + 21(−5)] = −979.

How many terms of 24, 21, 18, … give sum 78? n² − 17n + 52 = 0 gives n = 4 or 13. Both are valid because terms 5th to 13th add up to zero.

TV sets: a₃ = 600, a₇ = 700 gives d = 25, a = 550; production in the 10th year = 775; total in first 7 years = 4375.

Why the Sum Formula Works

Write the sum twice, once forwards and once backwards:

S = a + (a + d) + (a + 2d) + … + [a + (n − 1)d]

S = [a + (n − 1)d] + [a + (n − 2)d] + … + a

Adding term by term, every pair adds up to the same value 2a + (n − 1)d, and there are n pairs. So 2S = n[2a + (n − 1)d], which gives Sₙ = n/2[2a + (n − 1)d]. Since a + (n − 1)d is the last term l, this is also n/2(a + l). This is exactly Gauss's trick for 1 + 2 + … + 100.

Board Exam Focus

Questions on "which term", "is x a term", "nth term from the end" and "sum of n terms" appear every year.

Word problems (salary, savings, rows of plants, logs in a pile, prizes) — first identify a, d, n, then choose aₙ or Sₙ.

Always state the formula before substituting.

Formula Sheet

aₙ = a + (n − 1)d

d = aₙ − aₙ₋₁

Sₙ = n/2 [2a + (n − 1)d]

Sₙ = n/2 (a + l)

1 + 2 + … + n = n(n + 1)/2

aₙ = Sₙ − Sₙ₋₁

If a, b, c are in AP, then b = (a + c)/2 (b is the arithmetic mean)

Common Mistakes

Writing aₙ = a + nd (forgetting the n − 1).

Taking d = a₁ − a₂ instead of a₂ − a₁ (sign error in decreasing APs).

Accepting a fractional or negative n as a number of terms.

In "nth term from the end", counting the 11th from the end as the 14th instead of the 15th term (for 25 terms: 25 − 11 + 1 = 15).

Using Sₙ = n/2(a + l) when l is not the nth term.

💡 Key Learning Points

✓An AP is a list where each term is the previous term plus a fixed common difference d = aₖ₊₁ − aₖ.
✓The nth term is aₙ = a + (n − 1)d; x is a term only if the resulting n is a positive integer.
✓The sum of the first n terms is Sₙ = n/2[2a + (n − 1)d] = n/2(a + l).
✓The sum of the first n natural numbers is n(n + 1)/2 (Gauss's pairing idea).
✓Word problems become easy once a, d and n are identified correctly.

👨‍🏫 Teaching Tips

→Start with Reena's salary and the ladder rungs from the chapter opening; ask students to predict the 10th value before giving the formula.
→Let students discover d by subtracting consecutive terms for several lists, including one that is not an AP.
→Tell the Gauss story and have students pair 1 + 100, 2 + 99, … to derive n(n + 1)/2 themselves.
→Use Example 4 (−81 and 0 in 21, 18, 15, …) to show how n must be a positive integer.
→Discuss Example 13 where both n = 4 and n = 13 give sum 78, and ask why the terms 5th to 13th cancel.
→Give a 'build a word problem' activity: groups write an AP story (savings, seats in rows) and exchange for solving.
→Drill the difference between 'nth term' and 'sum of n terms' with quick oral questions.

📋 Assessment Questions

Find the 20th term of 3, 7, 11, … (Answer: 79).
Is 301 a term of 5, 11, 17, …? Justify.
Find the sum of the first 22 terms of 8, 3, −2, ….
How many two-digit numbers are divisible by 3?
A factory makes 600 units in year 3 and 700 in year 7 with uniform increase; find the output in year 10.

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