- Maths
- Classes 9-10
- 5 min read
Arithmetic progressions
Recognise a sequence with a constant difference, find any term with a + (n − 1)d and add up n terms with the sum formula.
Let us begin
The big idea
Stairs rise by the same height every step, and a savings plan can grow by the same amount every month. Sequences like these, where each term is the previous one plus a fixed number, are arithmetic progressions.
An arithmetic progression (AP) is a list of numbers in which each term is obtained by adding the same number d, called the common difference, to the term before. In 3, 7, 11, 15, ... the first term a is 3 and d is 4. The difference d can be negative, as in 10, 7, 4, 1, ... where d = −3.
The nth term is a + (n − 1)d. You add d once to get the second term, twice to get the third term, so to reach the nth term you add d exactly n − 1 times.
The sum of the first n terms is S = n/2 × [2a + (n − 1)d]. If you know the last term l, the same sum is S = n/2 × (a + l): the number of terms times the average of the first and last term.
To test whether a list is an AP, subtract neighbouring terms. If every difference is the same, it is an AP. The list 2, 4, 8, 16 is not an AP because the differences 2, 4, 8 keep changing.
Worked examples
See how it works
The 20th term of 3, 7, 11, ...
Here a = 3 and d = 4. The 20th term is 3 + (20 − 1) × 4 = 3 + 76 = 79.
The sum of those 20 terms
The first term is 3 and the 20th is 79, so S = 20/2 × (3 + 79) = 10 × 82 = 820.
Which term of 5, 8, 11, ... is 50?
We need 5 + (n − 1) × 3 = 50, so (n − 1) × 3 = 45, n − 1 = 15 and n = 16. So 50 is the 16th term.
A savings plan
Riya saves 100 rupees in January and 20 rupees more each month than the month before. In 12 months she saves S = 12/2 × [2 × 100 + 11 × 20] = 6 × (200 + 220) = 6 × 420 = 2520 rupees.
Try it at home
Build an AP with coins
You will need
- About 30 coins or buttons
- Paper and a pencil
What to do
- Make a row of 2 coins, then a row of 5, then 8, then 11. Each row has 3 more coins than the one before.
- Write the sequence: 2, 5, 8, 11. Name a and d. Predict the next two rows and then build them to check.
- Use the formula to find the 10th row without building it: 2 + 9 × 3 = 29.
- Count all the coins in the first four rows: 2 + 5 + 8 + 11 = 26. Now use the sum formula: 4/2 × (2 + 11) = 2 × 13 = 26. The formula matches the count.
Ask a grown-up: Keep small coins and buttons away from young children.
Practice sheet
Check what you learned
Practice sheet: Arithmetic progressions (Classes 9-10, Maths)
- What is the next term of 2, 5, 8, 11, ...?
- What is the common difference of 10, 7, 4, 1, ...?
- Find the 15th term of 3, 7, 11, ...
- Is 2, 4, 8, 16 an AP? Why or why not?
- Write the formula for the nth term of an AP.
- Find the sum of the first 10 terms of 1, 3, 5, 7, ...
- Which term of 7, 10, 13, ... is 61?
- Find the sum 1 + 2 + 3 + ... + 100.
- Three terms of an AP add up to 15. What is the middle term?
- An AP has first term 5 and common difference 2. Find the sum of its first 20 terms.
Answer key
Answers: Arithmetic progressions
- 14, because d = 3.
- −3, since 7 − 10 = −3.
- 3 + 14 × 4 = 59.
- No. The differences are 2, 4 and 8, not the same each time.
- nth term = a + (n − 1)d.
- S = 10/2 × [2 × 1 + 9 × 2] = 5 × 20 = 100.
- 7 + (n − 1) × 3 = 61 gives n − 1 = 18, so n = 19: the 19th term.
- 100/2 × (1 + 100) = 50 × 101 = 5050.
- 5, because three terms in an AP are (m − d), m and (m + d), which add to 3m.
- S = 20/2 × [2 × 5 + 19 × 2] = 10 × 48 = 480.