Number Patterns

Progressions & Series

Three patterns that build the backbone of sequences — arithmetic, geometric, and harmonic. Explore their history, formulae, and examples, then try the interactive calculators.

// number patterns → AP

Arithmetic Progression

A sequence where each term is obtained by adding the same fixed number to the previous term. Simple, powerful, and everywhere.

Origin & History

Where did it come from?

Arithmetic progressions trace back to ancient Egypt and Babylon (c. 1800 BCE), where scribes used them to divide grain fairly and plan irrigation calendars. The famous Rhind Mathematical Papyrus contains AP problems solved nearly 4,000 years ago.

Greek mathematicians formalised the idea — Euclid (c. 300 BCE) discussed arithmetic sequences in Elements. But the most legendary story belongs to Carl Friedrich Gauss (1777–1855), who as a child summed 1 + 2 + … + 100 = 5050 in seconds by pairing the ends: 1+100, 2+99, … — the very insight behind today's sum formula.

Introduction

The core idea

Imagine saving ₹100 on Monday, ₹200 on Tuesday, ₹300 on Wednesday… You add a fixed amount (₹100) each day. That's an AP.

The fixed amount added each time is called the common difference (d). The first term is a.

Key Formulae

nth Term & Sum

Arithmetic Progression — a, a+d, a+2d, …

nth term (Tₙ)Tₙ = a + (n − 1) × d
Sum of n terms (Sₙ)Sₙ = (n / 2) × [2a + (n − 1) × d]
Alternate sum formSₙ = (n / 2) × (first + last term)
Common differenced = Tₙ − Tₙ₋₁ (constant for all n)
Find n (given Tₙ)n = (Tₙ − a) / d + 1
Find n (given Sₙ)solve: n² × d/2 + n × (a − d/2) − Sₙ = 0
Simple Example

Step-by-step walkthrough

📘 Simple Example
AP: 3, 7, 11, 15, … — find the 10th term and sum of first 10 terms.
  • ①Identify: a = 3, d = 7 − 3 = 4
  • ②T₁₀ = 3 + (10 − 1) × 4 = 3 + 36 = 39
  • ③S₁₀ = (10/2) × [2×3 + 9×4] = 5 × 42 = 210
  • ④Check: (n/2)(first + last) = 5 × (3 + 39) = 210 ✓
T₁₀ = 39  |  S₁₀ = 210
Complex Example

A harder problem

🔑 Complex Example
The 5th term of an AP is 23 and its 12th term is 51. Find the AP and the sum of its first 20 terms.
  • ①T₅ = a + 4d = 23 … (i)
  • ②T₁₂ = a + 11d = 51 … (ii)
  • ③Subtract (i) from (ii): 7d = 28 → d = 4
  • ④From (i): a = 23 − 16 = 7
  • ⑤AP: 7, 11, 15, 19, 23, …
  • ⑥S₂₀ = (20/2)[2×7 + 19×4] = 10 × 90 = 900
AP: 7, 11, 15, …  |  S₂₀ = 900
Try it yourself

AP Calculator

Compute any AP — forward or reverse