In an arithmetic progression, each term is found by adding the same number d to the one before. The nth term is T_n = a + (n − 1)d, and the sum of the first n terms is S_n = n/2 [2a + (n − 1)d].
This lesson is part of progressions. If you are unsure which type of sequence you have, read the test on that page first.
What do a, d and n stand for?
- a is the first term.
- d is the common difference, the same amount added each step.
- n is the position of the term you want.
Write these three down before using any formula. Most wrong answers start with a wrong a, d or n.
Worked example: a term and a sum
Here is an original example. An arithmetic progression has first term 7 and common difference 4. Find T₂₀ and S₂₀.
T₂₀. T₂₀ = 7 + (20 − 1) × 4 = 7 + 76 = 83.
S₂₀. Use the first and last terms: S₂₀ = 20/2 × (7 + 83) = 10 × 90 = 900.
Check with the other formula: S₂₀ = 20/2 × [2(7) + 19(4)] = 10 × (14 + 76) = 900. Both agree.
The mistake that costs marks
The common slip is to multiply d by n instead of n − 1, so T₂₀ = 7 + 20 × 4 = 87. The working looks tidy and the answer is close, so it is hard to spot.
| Step | Wrong | Right |
|---|---|---|
| Number of d’s added | 20 | 19 |
| T₂₀ | 7 + 80 | 7 + 76 |
| Result | 87 | 83 |
A quick check: T₁ = a. If you put n = 1 into your formula and do not get a back, the formula is wrong. With n × d, you would get 7 + 4 = 11 for T₁, which is impossible.
Finding a and d from two terms
Some questions give two terms and ask for the progression. Let T₅ = 23 and T₁₂ = 58.
- Subtract to cancel a: T₁₂ − T₅ = 7d, so 58 − 23 = 7d.
- Solve: 7d = 35, so d = 5.
- Substitute back: T₅ = a + 4d = 23, so a + 20 = 23, and a = 3.
- Check with the other term: T₁₂ = 3 + 11 × 5 = 58. It matches.
The gap between the two term numbers (12 − 5 = 7) is how many d’s separate them. That idea makes the subtraction step easy to remember.
A decreasing progression
Take a = 12 and d = −3. Find S₁₀, and find which term first equals zero.
S₁₀ = 10/2 × [2(12) + 9(−3)] = 5 × (24 − 27) = 5 × (−3) = −15.
For the zero term: 12 + (n − 1)(−3) = 0, so n − 1 = 4 and n = 5. The fifth term is 0. After that the terms are negative, which is why the sum turns negative.
Check yourself
The first term of an arithmetic progression is 2 and the common difference is 6. Find the smallest n for which T_n exceeds 100.
Answer
T_n = 2 + (n − 1) × 6 = 6n − 4.
Solve 6n − 4 > 100, so 6n > 104 and n > 17.33.
The smallest whole n is 18.
Check: T₁₇ = 6(17) − 4 = 98, which is not above 100, and T₁₈ = 104, which is.
What to study next
The geometric version multiplies instead of adds, and its formulas have a power. Continue with terms and sums of geometric progressions, then test both types in the practice set.
If you want a teacher to check your a, d and n on new questions, see online one-to-one Additional Mathematics tuition.