CBSE Class 10 Maths Chapter 5 Notes — Arithmetic Progressions

What is an Arithmetic Progression?

A sequence a₁, a₂, a₃, ... is an Arithmetic Progression (AP) if the difference between consecutive terms is constant.

Common difference (d) = a₂ − a₁ = a₃ − a₂ = ...

Examples:

• 2, 5, 8, 11, 14, ... → d = 3 (increasing AP)
• 10, 7, 4, 1, −2, ... → d = −3 (decreasing AP)
• 3, 3, 3, 3, ... → d = 0 (constant AP)

Is a sequence an AP? Check if consecutive differences are all equal.

• 1, 4, 9, 16, ... → differences: 3, 5, 7, ... (not equal) → NOT an AP

nth Term of an AP

$$a_n = a + (n-1)d$$

Where:

• a = first term
• d = common difference
• n = position of the term
• aₙ = nth term

Worked examples:

Q1. Find the 20th term of AP: 3, 7, 11, 15, ...

a = 3, d = 4, n = 20

a₂₀ = 3 + (20−1) × 4 = 3 + 76 = 79

Q2. Which term of AP: 21, 18, 15, ... is −81?

a = 21, d = −3, aₙ = −81

−81 = 21 + (n−1)(−3) → −102 = (n−1)(−3) → n−1 = 34 → n = 35

Q3. How many 3-digit numbers are divisible by 7?

First 3-digit number divisible by 7: 105; Last: 994

AP: 105, 112, 119, ..., 994 → a = 105, d = 7, aₙ = 994

994 = 105 + (n−1)×7 → 889 = 7(n−1) → n−1 = 127 → n = 128

Sum of First n Terms

$$S_n = \frac{n}{2}[2a + (n-1)d]$$

When last term l is known: $$S_n = \frac{n}{2}(a + l)$$

Worked examples:

Q1. Find sum of first 20 terms of AP: 1, 3, 5, 7, ...

a = 1, d = 2, n = 20

S₂₀ = (20/2)[2×1 + 19×2] = 10[2 + 38] = 10 × 40 = 400

(This equals 20² — sum of first n odd numbers = n²)

Q2. Find sum of AP: 7, 10.5, 14, ..., 84

a = 7, d = 3.5, l = 84

First find n: 84 = 7 + (n−1)×3.5 → 77 = 3.5(n−1) → n = 23

Sₙ = (23/2)(7 + 84) = (23/2)(91) = 1046.5

Q3. If S₇ = 49 and S₁₇ = 289, find S_n.

S₇ = (7/2)(2a + 6d) = 49 → 2a + 6d = 14 ... (i) S₁₇ = (17/2)(2a + 16d) = 289 → 2a + 16d = 34 ... (ii)

Subtract (i) from (ii): 10d = 20 → d = 2; from (i): 2a = 2 → a = 1

Sₙ = (n/2)(2 + (n−1)×2) = (n/2)(2n) = n²

Finding aₙ from Sₙ

$$a_n = S_n - S_{n-1} \quad (n \geq 2)$$ $$a_1 = S_1$$

Worked example: Given Sₙ = 3n² + 5n. Find the nth term.

aₙ = Sₙ − Sₙ₋₁ = (3n² + 5n) − (3(n−1)² + 5(n−1)) = 3n² + 5n − 3n² + 6n − 3 − 5n + 5 = 6n + 2

Verify: a₁ = S₁ = 3 + 5 = 8. Formula: 6(1)+2 = 8 ✓

Key Properties of AP

• If a, b, c are in AP, then b − a = c − b, i.e., 2b = a + c (b is the arithmetic mean)
• Three terms in AP: Take them as a−d, a, a+d (sum = 3a)
• Four terms in AP: Take them as a−3d, a−d, a+d, a+3d (sum = 4a)
• If each term of AP is multiplied/divided by a constant, the result is also an AP
• Sum of terms equidistant from beginning and end = a + l (constant)

Board Exam Tips

• 6–8 marks from this chapter — appears every year
• Always identify a, d, and n clearly before substituting
• Finding which term is zero: Set aₙ = 0 and solve for n
• Sum formula with last term Sₙ = n/2(a+l) saves time when last term is given
• aₙ from Sₙ type questions: remember the formula aₙ = Sₙ − Sₙ₋₁; check a₁ = S₁ separately
• Word problems: "first, second, third installment..." → recognize as AP
• Three consecutive terms trick: use a−d, a, a+d to avoid extra algebra

📚 RD Sharma Mathematics Class 10 — AP Chapter with Extra Problems

📝 Oswaal CBSE Maths Question Bank — AP Board Questions