What is an arithmetic progression?
An arithmetic progression (AP) is a list of numbers in which each term is obtained by adding a fixed number to the previous term. That fixed number is the common difference, written d.
3, 7, 11, 15, … is an AP with first term a = 3 and d = 4.
20, 17, 14, 11, … is an AP with a = 20 and d = −3.
5, 5, 5, 5, … is an AP with d = 0.
To test whether a list is an AP, subtract each term from the next one. If every difference is the same, it is an AP. The list 1, 4, 9, 16, … is not an AP, because the differences are 3, 5, 7, …
The nth term
If the first term is a and the common difference is d, the terms are
a, a + d, a + 2d, a + 3d, …
Notice that the 4th term is a + 3d: the number of times d is added is always one less than the position. So the nth term (also called the general term) is
aₙ = a + (n − 1)d
Example. Find the 20th term of 7, 13, 19, …
Here a = 7 and d = 6, so a₂₀ = 7 + 19 × 6 = 7 + 114 = 121.
Example. Which term of the AP 3, 8, 13, … is 78?
Set aₙ = 78: 3 + (n − 1) × 5 = 78, so (n − 1) × 5 = 75 and n − 1 = 15. So 78 is the 16th term.
If solving gives a value of n that is not a positive whole number, the given number is not a term of the AP.
The nth term from the end
For an AP with last term l, counting backwards is just another AP with first term l and common difference −d. So the nth term from the end is l − (n − 1)d.
Sum of the first n terms
A well-known story says that the young Carl Friedrich Gauss added 1 + 2 + … + 100 by pairing terms from opposite ends: 1 + 100, 2 + 99, 3 + 98, … Each pair adds to 101, and there are 50 pairs, giving 5050.
The same pairing idea works for any AP and gives the sum of the first n terms:
Sₙ = n/2 × [2a + (n − 1)d]
If you know the last term l, an equivalent and often quicker form is
Sₙ = n/2 × (a + l)
Example. Find the sum of the first 25 terms of 2, 6, 10, …
a = 2, d = 4, n = 25: S₂₅ = 25/2 × [4 + 24 × 4] = 25/2 × 100 = 1250.
Example. Find the sum of all two-digit multiples of 7.
They are 14, 21, …, 98. Here a = 14, d = 7, l = 98. From 98 = 14 + (n − 1) × 7 we get n = 13. So S = 13/2 × (14 + 98) = 13 × 56 = 728.
Connecting aₙ and Sₙ
Because Sₙ is the sum of the first n terms, the nth term can be recovered as
aₙ = Sₙ − Sₙ₋₁
This is useful when a question gives you a formula for the sum and asks for a particular term.
Common mistakes to avoid
Using n instead of (n − 1) in the nth-term formula.
Getting the sign of d wrong for decreasing APs. Always compute d = a₂ − a₁.
Forgetting to check that n comes out as a positive integer.
Quick summary
Quantity | Formula |
|---|---|
Common difference | d = aₙ₊₁ − aₙ |
nth term | aₙ = a + (n − 1)d |
Sum of n terms | Sₙ = n/2 [2a + (n − 1)d] = n/2 (a + l) |
nth term from sums | aₙ = Sₙ − Sₙ₋₁ |