The nth term of an AP
The nth term of an AP is aₙ = a + (n − 1)d. Use it to find any term, a term's position, or the number of terms.
Where the formula comes from
Look at the pattern: a₁ = a, a₂ = a + d, a₃ = a + 2d, a₄ = a + 3d. Each time, the number of d's is one less than the position. So the nth term is aₙ = a + (n − 1)d. This is also called the general term. If an AP has m terms, its last term is l = a + (m − 1)d.
Kinds of questions and what to do
| You are asked | What to do |
|---|---|
| Find a given term, e.g. the 20th term | Put n = 20 in aₙ = a + (n − 1)d. |
| Which term is a given number? | Put aₙ equal to that number and solve for n. n must be a positive whole number. |
| How many terms are there? | Put aₙ equal to the last term and solve for n. |
| Find the AP from two of its terms | Write each term as a + (n − 1)d, subtract the two equations to get d, then find a. |
| nth term from the end | Use l − (n − 1)d, where l is the last term. |
| Three numbers in AP with a given sum | Take them as a − d, a, a + d (here a is the middle term). The d's cancel when you add them. |
Example 1: a term and a position
For the AP 25, 21, 17, ..., find (i) the 25th term and (ii) which term is −79.
- Here a = 25 and d = 21 − 25 = −4.
- Formula: aₙ = a + (n − 1)d
- (i) a₂₅ = 25 + (25 − 1) × (−4) = 25 + 24 × (−4)
- a₂₅ = 25 − 96 = −71
- (ii) Put aₙ = −79: 25 + (n − 1) × (−4) = −79
- (n − 1) × (−4) = −79 − 25 = −104
- n − 1 = 26, so n = 27
(i) a₂₅ = −71 (ii) −79 is the 27th term.
Example 2: the AP from two terms
The 4th term of an AP is 17 and its 10th term is 41. Find the AP and its 30th term.
- a₄ = a + 3d = 17 ... (1)
- a₁₀ = a + 9d = 41 ... (2)
- Subtract (1) from (2): 6d = 24, so d = 4
- Put d = 4 in (1): a + 12 = 17, so a = 5
- The AP is 5, 9, 13, 17, ...
- a₃₀ = 5 + 29 × 4 = 5 + 116 = 121
The AP is 5, 9, 13, 17, ... and its 30th term is 121.
Example 3: how many terms?
How many two-digit numbers are divisible by 7?
- The two-digit multiples of 7 are 14, 21, 28, ..., 98.
- This is an AP with a = 14, d = 7 and last term 98.
- Formula: aₙ = a + (n − 1)d, so 14 + (n − 1) × 7 = 98
- (n − 1) × 7 = 84
- n − 1 = 12, so n = 13
13 two-digit numbers are divisible by 7.
If a school fee rises by the same amount every year, aₙ = a + (n − 1)d gives the fee in any year. You need not list every year.
1 marks: “The 12th term of the AP −4, −3/2, 1, ... is:” d = −3/2 − (−4) = 5/2. a₁₂ = −4 + 11 × 5/2 = −4 + 55/2 = 47/2.
2 marks: “Which term of the AP 4, 9, 14, 19, ... is 104?” a = 4, d = 5. 4 + (n − 1) × 5 = 104 gives n − 1 = 20, so n = 21: the 21st term.
3 marks: “The 6th term of an AP is 27 and its 12th term is 57. Find the AP.” a + 5d = 27 and a + 11d = 57. Subtract: 6d = 30, so d = 5 and a = 2. The AP is 2, 7, 12, 17, ...
n − 1 jumps: to reach the nth term, start at a and take (n − 1) jumps of size d.