Finding the constant difference, writing the general term Tₙ = dn + c, calculating any term, and working backwards to find which term has a given value.
A sequence is linear (or arithmetic) when you add the same amount each time. That amount is the constant difference, d.
1The general term
d = constant difference
c = the term BEFORE the first one (that is, T₀)
Because Tₙ = dn + c is the equation of a straight line, the constant difference d is exactly the gradient. That is why the plotted points are collinear.
- 1d = 8 − 5 = 3. Check: 11 − 8 = 3 and 14 − 11 = 3. ✓Always check the difference is the same all the way.
- 2Tₙ = 3n + c.Substitute d = 3.
- 3T₁ = 5, so 3(1) + c = 5.Use the first term to find c.
- 4c = 2, so Tₙ = 3n + 2.Check T₃: 3(3) + 2 = 11. ✓
c is not the first term. For 5 ; 8 ; 11 the first term is 5 but c = 2 — it is the value the pattern would have had at n = 0.
2Finding a specific term
- 1Tₙ = 3n + 2.
- 2T₅₀ = 3(50) + 2.Substitute n = 50.
- 3= 152.
3Working backwards
- 13n + 2 = 47.Set the general term equal to the value.
- 23n = 45.
- 3n = 15, so 47 is the 15th term.Check: 3(15) + 2 = 47. ✓
- 13n + 2 = 100 → 3n = 98.
- 2n = 98/3 = 32,67…
- 3n is not a whole number, so 100 is not a term of this sequence.Term numbers must be positive integers — say this in words.
4Finding the pattern from two terms that are not next to each other
Sometimes the terms you are given are not consecutive. The difference between two terms is d multiplied by the number of steps between them.
- 1From T₃ to T₇ is 4 steps, so 4d = 27 − 11 = 16 and d = 4.Count steps, not terms.
- 2Tₙ = 4n + c and T₃ = 11: 4(3) + c = 11, so c = −1.
- 3Tₙ = 4n − 1.Check T₇: 4(7) − 1 = 27. ✓
5Decreasing linear patterns
- 1d = 27 − 31 = −4.A decreasing pattern has a negative constant difference.
- 2Tₙ = −4n + c; T₁ = 31 gives −4 + c = 31, so c = 35 and Tₙ = −4n + 35.
- 3−4n + 35 < 0 → n > 8,75, so the first negative term is T₉ = −4(9) + 35 = −1.n must be a natural number, so take the next whole number after 8,75.
6Linear patterns in context
- 1The costs for 1, 2 and 3 km are R24 ; R33 ; R42 ; …, so the constant difference is 9.
- 2Tₙ = 9n + 15.Here c = 15 is the starting fee: the cost at n = 0.
- 39n + 15 = 222 → 9n = 207 → n = 23 km.
In a context, c usually means something: it is the amount before anything has happened (n = 0) — a starting fee, an opening balance or a starting height.
Practice exercises
Work each one out, then click to reveal the answer.
- 1Write down the constant difference of 5 ; 8 ; 11 ; 14 ; …
Show answer ▾Hide answer ▴
d = 3 - 2Determine the general term Tₙ of 5 ; 8 ; 11 ; 14 ; …
Show answer ▾Hide answer ▴
d = 3, T₁ = 5 → c = 2 → Tₙ = 3n + 2 - 3Calculate T₅₀ for the sequence 5 ; 8 ; 11 ; …
Show answer ▾Hide answer ▴
3(50) + 2 = 152 - 4Which term of 5 ; 8 ; 11 ; … is equal to 47?
Show answer ▾Hide answer ▴
3n + 2 = 47 → n = 15 - 5Is 100 a term of the sequence 5 ; 8 ; 11 ; …? Justify.
Show answer ▾Hide answer ▴
3n = 98 gives n = 32,67 which is not a whole number, so no. - 6Determine the general term of 7 ; 11 ; 15 ; 19 ; …
Show answer ▾Hide answer ▴
d = 4, c = 3 → Tₙ = 4n + 3 - 7Determine the general term of 20 ; 17 ; 14 ; 11 ; …
Show answer ▾Hide answer ▴
d = −3, c = 23 → Tₙ = −3n + 23 - 8Explain why a linear sequence plots as a straight line.
Show answer ▾Hide answer ▴
Because Tₙ = dn + c is the equation of a straight line, with the constant difference as the gradient.
Quick Quiz
5 quick questions on what you just read. Take it when you feel ready.
Now practise it
Download Grade 10 past papers and worksheets on this topic.