Revision notes
Spot the pattern, find any term with Tₙ = a + (n − 1)d, and add up many terms fast with Sₙ.
Chidi saves ₦500 this week. Every week after, he saves ₦200 more than the week before: ₦500, ₦700, ₦900, … How much will he save in week 10? And how much altogether? You could add them one by one — or use a pattern called an arithmetic progression.
A sequence is a list of numbers in order. Each number is a term: T₁, T₂, T₃, …
In an arithmetic progression, you ADD the same number every time. That number is the common difference, d.
The first term is called a. In 500, 700, 900, … we have a = 500 and d = 200.
🔑 Key idea: d = any term − the term before it.
To reach the 2nd term you make 1 jump. To reach the 3rd, 2 jumps. To reach the nth term, (n − 1) jumps of size d.
So Tₙ = a + (n − 1)d.
🔑 Key idea: Tₙ = a + (n − 1)d
Find T₁₀ of 3, 7, 11, …
Step 1: Find a and d
a = 3
d = 7 − 3 = 4
Step 2: Use Tn = a + (n − 1)d
T10 = 3 + (10 − 1) × 4
= 3 + 9 × 4
= 39
Answer: T₁₀ = 39
Legend says a young mathematician, Gauss, added 1 + 2 + … + 100 in seconds: he paired 1 + 100, 2 + 99, 3 + 98… — 50 pairs of 101 = 5050.
The same trick gives the sum of the first n terms: Sₙ = n/2 × [2a + (n − 1)d].
🔑 Key idea: Sₙ = n/2 [2a + (n − 1)d]
Sum of the first 20 terms of 2, 5, 8, …
Step 1: Identify a, d and n
a = 2, d = 3, n = 20
Step 2: Work out the bracket
2a + (n − 1)d = 2 × 2 + 19 × 3 = 61
Step 3: Multiply by n/2
S20 = 202 × 61 = 610
Answer: S₂₀ = 610
Ready to check you've got it? Scroll up and work through the lesson: the interactive visuals, practice questions and mastery check are all there.