POP

G8.5 Quadratic Sequences

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01 — Explanation

How do I recognise a quadratic sequence?

A linear sequence has a constant first difference; a quadratic sequence $T_n=an^2+bn+c$ has a constant second difference instead. Find it by differencing twice:

Sequence: 5, 7, 11, 17, 25 → 1st diff: 2, 4, 6, 8 → 2nd diff: 2, 2, 2 (constant → quadratic confirmed).

This topic is Higher Tier only — Foundation candidates are not expected to find the $n$ th term of a quadratic sequence.

02 — Explanation

How do I find the formula quickly (the algebraic method)?

Rather than the slower "subtract $an^2$ " approach, derive $a$ , $b$ , $c$ directly from three values you already found while differencing:

2a = \text{constant 2nd difference} \qquad 3a+b = \text{first 1st difference} \qquad a+b+c = \text{first term}

Solve these in order — $a$ first, then $b$ , then $c$ .

Worked example: Find the $n$ th term of 5, 7, 11, 17,...

1st diff: 2, 4, 6. 2nd diff: 2, 2.

2a=2 \implies a=1 \qquad 3(1)+b=2 \implies b=-1 \qquad 1+(-1)+c=5 \implies c=5

T_n = n^2-n+5

Exam Tip

This method is faster and more reliable than the "subtraction" technique, but only if you have the three equations memorised correctly — the most common slip is forgetting it's $3a+b$ (not $a+b$ ) for the first difference.

03 — Explanation

What if the second difference is negative?

The method is identical — just be careful subtracting negatives. For 19, 15, 9, 1: 1st diff −4,−6,−8; 2nd diff −2,−2. So $2a=-2\implies a=-1$ ; $3(-1)+b=-4\implies b=-1$ ; $-1-1+c=19\implies c=21$ . $T_n=-n^2-n+21$ .

04 — Explanation

How do I verify my formula is correct?

Substitute a known position back in. For $T_n=n^2-n+5$ , check $n=3$ : $9-3+5=11$ — matches the 3rd term. Always do this; it catches sign errors in seconds.

05 — Explanation

How do I answer "is [value] a term in this sequence?"

Set $T_n$ equal to the value and solve for $n$ . The critical rule: $n$ must come out as a positive integer — if it doesn't, the value is not in the sequence.

Worked example: Is 100 a term of $T_n=n^2+n+4$ ?

n^2+n+4=100 \implies n^2+n-96=0

This doesn't factorise cleanly; the quadratic formula gives $n=\frac{-1\pm\sqrt{385}}{2}$ , and $\sqrt{385}\approx19.6$ is not an integer. So 100 is not a term.

06 — Explanation

How do I find which term has a given value?

Same method, but set up so the quadratic does factorise nicely — then reject whichever solution for $n$ isn't a valid positive integer.

Worked example: A sequence starts 6, 10, 16, 24,... (a) Show the $n$ th term is $n^2+n+4$ . (b) Hence find the term with value 136.

(a) 1st diff: 4,6,8. 2nd diff: 2,2. So $2a=2\Rightarrow a=1$ ; $3(1)+b=4\Rightarrow b=1$ ; $1+1+c=6\Rightarrow c=4$ . This gives $n^2+n+4$ , as required.

(b) Set $n^2+n+4=136 \implies n^2+n-132=0$ . Factorise: numbers multiplying to $-132$ and summing to $1$ are $12$ and $-11$ , so $(n+12)(n-11)=0$ , giving $n=-12$ or $n=11$ .

Since $n$ represents a position, it cannot be negative — reject $n=-12$ . The 11th term has value 136.

Common Mistake

Writing both $n=-12$ and $n=11$ as the final answer. The question asks "find the term" — a single, contextual answer is required; showing you've rejected the invalid solution is itself worth marks.

07 — Worked example

Worked exam-style question

Work out the formula for the $n$ th term of 2, 10, 24, 44,...

1st diff: 8, 14, 20. 2nd diff: 6, 6.

2a=6 \implies a=3 \qquad 3(3)+b=8 \implies b=-1 \qquad 3+(-1)+c=2 \implies c=0

T_n = 3n^2-n

Verify with $n=2$ : $3(4)-2=10$ ✓ — matches the 2nd term.