Quadratic Number Patterns (Grade 11 & 12)

Recognising a quadratic pattern from its constant second difference, finding Tₙ = an² + bn + c, and answering the standard NSC follow-ups about which term equals a given value.

A sequence is quadratic when the second differences are constant. Paper 1 opens with one of these almost every year.

1Spotting it: the difference table

The sequence 3 ; 8 ; 17 ; 30 ; 47 — second differences are all 4.

Subtract each term from the next to get the first differences, then do it again for the second differences. Constant second difference ⇒ quadratic.

2The three shortcuts

Tₙ = an² + bn + c

2a = second difference
3a + b = first of the first differences
a + b + c = T₁

💡

Learn those three lines. They turn every quadratic-pattern question into three quick substitutions — no simultaneous equations needed.

Worked ExampleFind Tₙ for 3 ; 8 ; 17 ; 30 ; 47

1
Second difference = 4, so 2a = 4 and a = 2.
2
First of the first differences = 5, so 3a + b = 5 → 6 + b = 5 → b = −1.
Substitute a = 2.
3
T₁ = 3, so a + b + c = 3 → 2 − 1 + c = 3 → c = 2.
4
Tₙ = 2n² − n + 2.
5
Check with n = 4: 2(16) − 4 + 2 = 30. ✓
ALWAYS test one known term — it catches every sign slip.

⚠️

Do not use the first difference between T₁ and T₂ as '2a'. The second difference is the one that is constant; the first differences keep changing.

3Which term equals a given value?

Worked ExampleWhich term of Tₙ = 2n² − n + 2 equals 122?

1
2n² − n + 2 = 122.
Set the formula equal to the value.
2
2n² − n − 120 = 0.
Standard form.
3
(2n + 15)(n − 8) = 0.
Factorise.
4
n = 8 (reject n = −7,5).
n must be a positive whole number — say so, marks are given for it.

4Finding a missing value

Worked Example1 ; p ; 11 ; … is quadratic with first differences 3 ; 7 ; …

1
First difference: p − 1 = 3.
The gap from T₁ to T₂.
2
p = 4.
3
Check: 11 − 4 = 7. ✓
Matches the second given first-difference.

Practice exercises

Work each one out, then click to reveal the answer.

📄 Worksheet ✅ Memo 📝 Exam paper 🗝️ Marking guide

1
Show that 3 ; 8 ; 17 ; 30 ; 47 is a quadratic pattern.
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First differences 5 ; 9 ; 13 ; 17. Second differences 4 ; 4 ; 4 — constant, so it is quadratic.
2
Determine Tₙ for 3 ; 8 ; 17 ; 30 ; 47
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2a = 4 → a = 2; 3a + b = 5 → b = −1; a+b+c = 3 → c = 2. Tₙ = 2n² − n + 2
3
Which term of 3 ; 8 ; 17 ; 30 ; 47 equals 122?
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2n² − n − 120 = 0 → (2n+15)(n−8) = 0 → n = 8 (reject n = −7,5)
4
Determine the general term of the pattern shown in the difference table below.
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2a = 2 → a = 1; 3a + b = 4 → b = 1; a+b+c = 1 → c = −1. Tₙ = n² + n − 1
5
1 ; p ; 11 ; … is a quadratic pattern whose first differences are 3 ; 7 ; … Determine p.
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p − 1 = 3, so p = 4
6
The second difference of a quadratic pattern is 6. Determine a.
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2a = 6, so a = 3
7
Calculate T₁₀ if Tₙ = 2n² − n + 2.
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2(100) − 10 + 2 = 192