What an arithmetic progression is
An arithmetic progression (AP) is a list where each term is the previous term plus a fixed number. That number is the common difference d.
What it means
A sequence is a list of numbers in a fixed order. Each number in it is a term. In an arithmetic progression (AP), you get each term by adding a fixed number to the term before it. This fixed number is the common difference, written d. The first term is written a (or a₁), the second a₂, and so on.
Examples and non-examples
| List | Differences | AP? |
|---|---|---|
| 2, 5, 8, 11, ... | 3, 3, 3 | Yes, d = 3 |
| 50, 45, 40, 35, ... | −5, −5, −5 | Yes, d = −5 |
| 7, 7, 7, 7, ... | 0, 0, 0 | Yes, d = 0 |
| −1.5, −1, −0.5, 0, ... | 0.5, 0.5, 0.5 | Yes, d = 0.5 |
| 1, 4, 9, 16, ... | 3, 5, 7 | No, the differences change |
| 3, 6, 12, 24, ... | 3, 6, 12 | No, each term is doubled, not added to |
Rules to remember
- A finite AP has a last term, e.g. the roll numbers 1, 2, 3, ..., 40 of a class.
- An infinite AP goes on forever, e.g. 3, 6, 9, 12, ... (the multiples of 3).
- Find d as a term minus the term just before it: d = a₂ − a₁, never a₁ − a₂.
- To show a list is an AP, find every difference you can and show they are equal. One pair is not enough.
- Three numbers a, b, c are in AP exactly when 2b = a + c.
Example 1: is it an AP?
Check whether √3, √12, √27, √48, ... is an AP. If it is, write the next term.
- Simplify each term: √12 = 2√3, √27 = 3√3, √48 = 4√3.
- So the list is √3, 2√3, 3√3, 4√3, ...
- a₂ − a₁ = 2√3 − √3 = √3
- a₃ − a₂ = 3√3 − 2√3 = √3 and a₄ − a₃ = 4√3 − 3√3 = √3
- All the differences are equal, so the list is an AP with d = √3.
- Next term = 4√3 + √3 = 5√3, which is √75.
Yes, it is an AP with d = √3. The next term is 5√3 (that is, √75).
Example 2: auto fares
An auto-rickshaw charges ₹ 30 for the first km and ₹ 15 for every extra km (example rates). Write the fares for 1, 2, 3 and 4 km. Do they form an AP?
- Fare for 1 km = ₹ 30
- Fare for 2 km = 30 + 15 = ₹ 45
- Fare for 3 km = 45 + 15 = ₹ 60
- Fare for 4 km = 60 + 15 = ₹ 75
- Differences: 45 − 30 = 15, 60 − 45 = 15, 75 − 60 = 15. All are equal.
The fares 30, 45, 60, 75, ... form an AP with a = 30 and d = 15.
APs are all around you. An auto fare rising by a fixed amount per km, or saving ₹ 10 more each week for Diwali, follows an AP.
1 marks: “The common difference of the AP 7/4, 5/4, 3/4, ... is:” d = 5/4 − 7/4 = −2/4 = −1/2.
2 marks: “Is 5, 10, 20, 40, ... an AP? Give a reason.” The differences are 5, 10 and 20. They are not equal, so it is not an AP.
2 marks: “For what value of k are 2k − 1, 7 and 3k three consecutive terms of an AP?” Use a₂ − a₁ = a₃ − a₂: 7 − (2k − 1) = 3k − 7, so 8 − 2k = 3k − 7 and 5k = 15. k = 3, giving 5, 7, 9.
Same jump every time = AP. Find d as later term minus earlier term.