Arithmetic Progression — Study Notes

Overview

Arithmetic Progression (AP) is one of the most predictable and high-scoring topics in the KAR TET Paper II Mathematics section. It forms the foundation for understanding sequences and series, and questions typically test your ability to identify patterns, find specific terms, and calculate sums.

This topic connects directly to real-life situations—calculating monthly savings, arranging seats in auditoriums, or understanding patterns in nature. From an exam perspective, expect 2–3 direct questions that reward formula memorisation and careful substitution. Mastering AP also builds your confidence for related concepts like geometric progression in higher studies.

The key to success here is understanding what makes a sequence "arithmetic" and knowing when to apply which formula. Most questions become straightforward once you identify the first term (a) and common difference (d).

Key Concepts

**Arithmetic Progression defined**: A sequence where each term after the first is obtained by adding a fixed number (common difference) to the previous term. Example: 3, 7, 11, 15, ... has common difference 4.

**Common difference (d)**: The constant value added to get the next term. Calculate as d = (any term) − (previous term). If d > 0, AP is increasing; if d < 0, AP is decreasing; if d = 0, all terms are equal.

**First term (a)**: The starting point of the AP. Every formula revolves around knowing 'a' and 'd'.

**General (nth) term**: Any term's position can be found without listing all previous terms. This is the core skill examiners test.

**Finite vs Infinite AP**: Finite AP has a definite last term (like 2, 5, 8, 11); infinite AP continues indefinitely (like 1, 2, 3, 4, ...).

**Three consecutive terms**: If three numbers are in AP, the middle term equals the average of the other two. So if a, b, c are in AP, then b = (a + c)/2, or equivalently, 2b = a + c.

**Sum of AP**: The total of all terms from the first to the nth term. This is frequently tested using both forms of the sum formula.

Formulas / Key Facts

**nth Term of AP** aₙ = a + (n − 1)d

aₙ = nth term, a = first term, n = position, d = common difference
Use when: finding any specific term or checking if a number belongs to the AP

**Sum of First n Terms (Form 1)** Sₙ = n/2 × [2a + (n − 1)d]

Use when: you know a, d, and n

**Sum of First n Terms (Form 2)** Sₙ = n/2 × (a + l)

l = last term
Use when: first and last terms are known (saves calculation)

**Common Difference** d = a₂ − a₁ = a₃ − a₂ = aₙ − aₙ₋₁

**Middle Term (odd number of terms)** Middle term = (n + 1)/2 th term

**Sum of First n Natural Numbers** 1 + 2 + 3 + ... + n = n(n + 1)/2

Special case of AP with a = 1, d = 1

**Relationship between Sₙ and aₙ** aₙ = Sₙ − Sₙ₋₁ (for n ≥ 2)

Worked Examples

**Example 1: Finding the nth term** *Find the 20th term of the AP: 7, 13, 19, 25, ...*

Step 1: Identify a and d

a = 7 (first term)
d = 13 − 7 = 6

Step 2: Apply the formula

a₂₀ = a + (n − 1)d
a₂₀ = 7 + (20 − 1) × 6
a₂₀ = 7 + 19 × 6
a₂₀ = 7 + 114 = **121**

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**Example 2: Finding the sum of first n terms** *Find the sum of first 15 terms of AP: 5, 9, 13, 17, ...*

Step 1: Identify values

a = 5, d = 9 − 5 = 4, n = 15

Step 2: Apply sum formula

S₁₅ = n/2 × [2a + (n − 1)d]
S₁₅ = 15/2 × [2(5) + (15 − 1)(4)]
S₁₅ = 15/2 × [10 + 56]
S₁₅ = 15/2 × 66
S₁₅ = 15 × 33 = **495**

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**Example 3: Finding how many terms** *How many terms of AP 24, 21, 18, ... must be taken so that their sum is 78?*

Step 1: Identify values

a = 24, d = 21 − 24 = −3, Sₙ = 78

Step 2: Use sum formula and solve for n

78 = n/2 × [2(24) + (n − 1)(−3)]
78 = n/2 × [48 − 3n + 3]
78 = n/2 × [51 − 3n]
156 = n(51 − 3n)
156 = 51n − 3n²
3n² − 51n + 156 = 0
n² − 17n + 52 = 0
(n − 4)(n − 13) = 0
n = 4 or n = 13

Both values are valid. **n = 4 or n = 13 terms**

(Verification: After 4 terms sum = 78; after 13 terms, negative terms reduce the total back to 78)

Common Mistakes

**Forgetting (n − 1) in the nth term formula** → Students write aₙ = a + nd instead of aₙ = a + (n − 1)d. Remember: the first term has zero additions of d.

**Sign error in common difference** → When AP is decreasing (like 20, 17, 14...), d is negative. Always calculate d = second term − first term, preserving the sign.

**Confusing nth term with sum** → The question asks for "15th term" but student calculates S₁₅. Read carefully: "term" means aₙ; "sum of terms" means Sₙ.

**Using wrong sum formula** → Students memorise only Sₙ = n/2(a + l) but don't know 'l'. When last term is unknown, use Sₙ = n/2[2a + (n − 1)d].

**Arithmetic errors with fractions** → When n is odd and you're dividing by 2, simplify carefully. For n = 15: 15/2 × 66 = 15 × 33, not 7.5 × 66.

Quick Reference

AP means: next term = previous term + d (constant)

nth term: aₙ = a + (n − 1)d

Sum: Sₙ = n/2 × [2a + (n − 1)d] OR n/2 × (first + last)

Common difference: d = any term − its previous term

If a, b, c in AP → 2b = a + c

To find n when Sₙ is given → form quadratic and solve

Arithmetic Progression