Arithmetic Progressions (AP)

Why This Chapter Matters

AP is tested in every CBSE Class 10 board exam — at minimum 5 marks, often more. The nth term and sum formulas are essential tools. Word problems involving APs test practical application and are commonly seen in the Long Answer section.

Prerequisites

•Sequences and series concept

•Basic algebra (substituting and solving linear equations)

Core Concepts

1. What is an Arithmetic Progression?

A sequence where the difference between consecutive terms is constant.

This constant difference is called the Common Difference (d).

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

Examples:

•2, 5, 8, 11, 14, ... → d = 3 (increasing AP)

•10, 7, 4, 1, −2, ... → d = −3 (decreasing AP)

•5, 5, 5, 5, ... → d = 0 (constant AP)

Checking if a sequence is AP: Subtract consecutive terms — if the difference is always the same, it's an AP.

2. nth Term Formula (General Term)

$$\boxed{a_n = a + (n-1)d}$$

Where:

•aₙ = nth term

•a = first term (a₁)

•d = common difference

•n = position of the term

Example: Find the 15th term of AP: 7, 13, 19, 25, ...

a = 7, d = 13 − 7 = 6, n = 15

a₁₅ = 7 + (15−1) × 6 = 7 + 84 = 91

Example: Which term of the AP 3, 8, 13, 18, ... is 78?

aₙ = 78 → 3 + (n−1)(5) = 78 → (n−1)(5) = 75 → n−1 = 15 → n = 16

3. Sum of First n Terms

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

Alternative form (when last term l = aₙ is known):

$$S_n = \frac{n}{2}[a + l]$$

Important relation: aₙ = Sₙ − Sₙ₋₁

Example: Find sum of first 20 terms of AP: −5, −8, −11, ...

a = −5, d = −3, n = 20

S₂₀ = (20/2)[2(−5) + 19(−3)] = 10[−10 − 57] = 10 × (−67) = −670

4. Properties of AP

•If a, b, c are in AP → b − a = c − b → 2b = a + c (b is the Arithmetic Mean of a and c)

•If a constant is added/subtracted/multiplied to each term of AP → resulting sequence is still AP

•If three terms in AP: take them as (a−d), a, (a+d) — makes calculation easier!

•If four terms in AP: take them as (a−3d), (a−d), (a+d), (a+3d)

Solved Examples

Example 1

Q: Find the sum of first 22 terms of AP in which d = 7 and the 22nd term is 149.

a₂₂ = a + 21d = 149 → a + 147 = 149 → a = 2

S₂₂ = (22/2)(a + a₂₂) = 11(2 + 149) = 11 × 151 = 1661

Example 2 — Word Problem

Q: The sum of 4th and 8th terms of an AP is 24, sum of 6th and 10th terms is 44. Find first three terms.

a₄ + a₈ = (a+3d) + (a+7d) = 2a + 10d = 24 → a + 5d = 12 ... (1)

a₆ + a₁₀ = (a+5d) + (a+9d) = 2a + 14d = 44 → a + 7d = 22 ... (2)

(2) − (1): 2d = 10 → d = 5

From (1): a = 12 − 25 = −13

First three terms: −13, −8, −3

Example 3 — Sum of Multiples

Q: Find the sum of all multiples of 7 between 1 and 500.

Multiples of 7 between 1 and 500: 7, 14, 21, ..., 497

a = 7, d = 7, last term = 497

497 = 7 + (n−1)7 → n = 71

S = (71/2)(7 + 497) = (71/2)(504) = 71 × 252 = 17892

PYQs

2023

Q: How many terms of AP 9, 17, 25, ... must be taken to get a sum of 636?

a=9, d=8, Sₙ=636

636 = (n/2)[18 + 8(n−1)] = n(4n+5)

4n² + 5n − 636 = 0 → n = 12 (taking positive value)

12 terms

2022

Q: Find nth term of AP: 7, 13, 19, ..., 205

a=7, d=6, aₙ=205 → 7 + (n−1)6 = 205 → n = 34

2021

Q: An AP consists of 50 terms. Third term is 12, last term is 106. Find 29th term.

a + 2d = 12 and a + 49d = 106 → 47d = 94 → d = 2, a = 8

a₂₉ = 8 + 28×2 = 64

2020

Q: The sum of first n terms of an AP is 3n² + 4n. Find nth term and AP.

Sₙ = 3n² + 4n

S₁ = 7 = a₁ (first term = 7)

S₂ = 12 + 8 = 20 → a₂ = 20 − 7 = 13

d = 13 − 7 = 6

aₙ = Sₙ − Sₙ₋₁ = 3n² + 4n − [3(n−1)² + 4(n−1)] = 6n + 1

MCQ Practice

Q1. The common difference of AP: 1/3, 5/3, 9/3, 13/3 is:

(A) 2/3 (B) 4/3 ✓ (C) 3 (D) 1

Q2. 11th term of AP −3, −1/2, 2, ... is:

(A) 22 ✓ (B) 28 (C) −38 (D) −48

[a=−3, d=2.5, a₁₁ = −3 + 10×2.5 = 22]

Q3 (Hard). If the sum of first p terms of AP equals sum of first q terms, find sum of first (p+q) terms:

Let Sₚ = Sq → (p/2)(2a+(p−1)d) = (q/2)(2a+(q−1)d)

After simplification: 2a + (p+q−1)d = 0 → S(p+q) = 0

Revision Notes

AP Key Formulas:
  aₙ = a + (n−1)d         ← nth term
  Sₙ = n/2 [2a + (n−1)d]  ← sum formula 1
  Sₙ = n/2 [a + l]         ← sum formula 2 (l = last term)
  aₙ = Sₙ − Sₙ₋₁          ← nth term from sum

Three terms in AP:  take (a−d), a, (a+d) → sum = 3a
Four terms in AP:   take (a−3d), (a−d), (a+d), (a+3d) → sum = 4a
Arithmetic Mean:    AM of a and b = (a+b)/2

Common Mistakes:

❌ Using n(n−1)/2 for sum instead of n/2[2a+(n−1)d]

❌ In "find how many terms" problems, rejecting n as non-integer but not checking

❌ Forgetting that if aₙ is negative for some n in a real-life problem, it's invalid

Arithmetic Progressions