Are quadratic sequences on the Foundation paper?

No. In Edexcel 1MA1 the nth term of a quadratic sequence is Higher tier only, under spec refs A24 and A25. Foundation candidates recognise square and triangular numbers and find the nth term of a linear sequence, but they are never asked to deduce a rule of the form an^2 + bn + c.

Why do you halve the second difference?

Because the second difference of an^2 + bn + c always works out as 2a. The bn + c part has a constant first difference, so it adds nothing at the second stage, and n^2 on its own has second difference 2. A measured second difference of 6 therefore means a = 3.

How many marks is a quadratic sequence question worth on Edexcel Higher?

Finding the nth term from a list of terms is usually 3 marks: one for the second difference and the an^2 it gives, one for the subtraction, one for the complete expression. Generating a term from a rule you are given is 1-2 marks, and the version that quotes the rule as an^2 + bn and supplies two terms runs to 4 or 5.

Is the second difference always 2?

No, that only happens when the coefficient of n^2 is 1, as it is for the square numbers. A second difference of 8 gives 4n^2, a second difference of 1 gives (1)/(2)n^2, which is how the triangular numbers 1, 3, 6, 10 behave, and a sequence that turns downwards has a negative second difference.

Quadratic Sequences - Edexcel GCSE Higher Maths

Typically 3-5 marks per paper when it appears · Spec 1MA1

A quadratic sequence is one whose nth term contains $n^2$ . The give-away is that the gaps between the terms are not constant but the gaps between those gaps are: 2, 9, 20, 35, 54 has first differences 7, 11, 15, 19 and a second difference of 4 all the way along. Edexcel files this under two references. A24 is recognising a quadratic sequence when you meet one, alongside arithmetic, geometric and Fibonacci type sequences, and A25 is deducing its nth term in the form $an^2 + bn + c$ . Both are Higher tier only in 1MA1, so a Foundation candidate is asked for the nth term of a linear sequence and taken no further.

The standard version hands you four or five terms and asks for an expression in terms of $n$ , worth 3 marks, and it can appear on any of the three papers. Nothing in the method needs a calculator, so Paper 1 is fair game and the terms there tend to stay small; Papers 2 and 3 are free to run the terms into the hundreds, because the arithmetic was never the point. Two variants come round often enough to rehearse separately. In one the rule is given and you generate a term, or work backwards from a term to its position. In the other the rule is quoted as $an^2 + bn$ with two terms supplied, which turns the question into a pair of simultaneous equations.

The three marks split into a method mark for the second difference and the $an^2$ it gives, a method mark for subtracting $an^2$ from the sequence, and an accuracy mark for the finished expression. Halving is where the first one disappears: a second difference of 6 means $3n^2$ , not $6n^2$ . The subtraction row decides the other two, and it has to be the sequence minus $an^2$ , in that order, every term, negatives included. Then substitute $n = 1$ and $n = 4$ back into your expression. If either misses, the error is in the row above, and you have twenty seconds to find it.

Worked example

Here are the first five terms of a quadratic sequence.

2, \quad 9, \quad 20, \quad 35, \quad 54

(a) Find an expression, in terms of

n

, for the nth term of this sequence. (3 marks) (b) Hence work out the 12th term of the sequence. (1 mark)

[4 marks]

Write the first differences underneath the terms: 7, 11, 15, 19. They are not constant, so take differences again: 4, 4, 4. The second difference is 4.
M1: constant second difference obtained, which is what confirms the sequence is quadratic
Half of 4 is 2, so the rule starts with $2n^2$ . Generating it gives 2, 8, 18, 32, 50.
M1: $a$ taken as half the second difference, with the terms of $2n^2$ written out
Subtract term by term, sequence minus $2n^2$ : 0, 1, 2, 3, 4. That is linear with nth term $n - 1$ , so the nth term of the sequence is $2n^2 + n - 1$ .
A1: fully correct expression; check at $n = 3$ , where $18 + 3 - 1 = 20$
Substitute $n = 12$ : $2(144) + 12 - 1 = 288 + 11 = 299$ .
B1: follow through from their part (a) expression, so a slip in (a) is not punished twice

Practice questions

These are original questions written in Edexcel 1MA1 Higher style. Every one can be done without a calculator, which is how Paper 1 sets them, though the harder questions use the sort of awkward numbers Papers 2 and 3 allow. Type an expression such as 2n^2 + n - 1 straight into the box and any equivalent form is accepted; where a question asks for $a$ , $b$ and $c$ separately you can enter a fraction like 1/2.

No account needed: the questions below are always free to check, plus 8 extra questions today. Your score so far updates as you go.

Question 1 Warm-up [1 mark]

The second difference of a quadratic sequence is 10. Write down the coefficient of $n^2$ in the nth term of this sequence.

Your answer (fractions like 1/2 are fine)
Question 2 Warm-up [2 marks]

The nth term of a sequence is $2n^2 - 1$ . Work out the 6th term.

Your answer (fractions like 1/2 are fine)
Question 3 Exam pace [3 marks]

Here are the first five terms of a quadratic sequence. $6, \quad 9, \quad 14, \quad 21, \quad 30$ Find an expression, in terms of $n$ , for the nth term of this sequence.

Your expression (equivalent forms accepted, e.g. 2(x+3) or 2x+6)
Question 4 Exam pace [2 marks]

The nth term of a sequence is $n^2 + 3n - 5$ . Work out the 12th term.

Your answer (fractions like 1/2 are fine)
Question 5 Exam pace [1 mark]

A quadratic sequence has a constant second difference of 6. Which of these could be the nth term of the sequence?

Answer options for question 5
$3n^2 + n - 2$ $6n^2 + n - 2$ $n^2 + 6n - 2$ $2n^2 + 6n - 2$
Question 6 Stretch [4 marks]

A stone is dropped from a tall cliff. It falls 5 metres in the first second, 20 metres in the first 2 seconds, 45 metres in the first 3 seconds and 80 metres in the first 4 seconds. The pattern continues in the same way. Work out how far the stone falls in the first 9 seconds. Give your answer in metres.

Your answer (fractions like 1/2 are fine)

Common mistakes examiners see

Using the second difference itself as the coefficient of $n^2$ , so a second difference of 6 produces $6n^2$ and every term comes out roughly double the size it should be.

Halve it. The second difference of $an^2 + bn + c$ is always $2a$ , so $a$ is half of what you measured: for 6, 19, 38, 63 the second difference is 6 and the rule starts $3n^2$ . The same slip wears a disguise on a decreasing sequence, where 9, 6, 1, $-6$ has second difference $-2$ and the rule starts $-n^2$ .
Stopping at the first differences, seeing them grow by 4 each time and writing a linear rule such as $4n + 2$ .

The moment the first differences fail to match, write a second row of differences before you choose a method. A constant first difference means $dn + k$ , a constant second difference means $an^2 + bn + c$ , and a constant ratio means the sequence is geometric.
Subtracting the wrong way round, $an^2$ minus the sequence, so every sign in the linear part flips and $2n^2 + n - 1$ is handed in as $2n^2 - n + 1$ .

Keep the original sequence as the top row and $an^2$ directly beneath it, then work down the columns. Substituting $n = 1$ into the finished expression and comparing with the first term catches a reversed sign in seconds.
Treating the leftover row as a constant when it is still changing, so 0, 1, 2, 3, 4 is read as 'add nothing' and the answer is left as $2n^2$ .

The row after subtraction is a linear sequence in its own right and only collapses to a single number when $b = 0$ . If it changes, find its nth term the way you would any arithmetic sequence and add the whole $bn + c$ on.
Squaring the coefficient along with $n$ when generating terms, so $3n^2$ at $n = 4$ is worked out as $(3 \times 4)^2 = 144$ rather than 48.

Square first, multiply second. Write the square numbers 1, 4, 9, 16, 25 as a row and then multiply each one by $a$ ; it fixes the order of operations and is faster than keying in five separate terms.

Frequently asked questions

Are quadratic sequences on the Foundation paper?: No. In Edexcel 1MA1 the nth term of a quadratic sequence is Higher tier only, under spec refs A24 and A25. Foundation candidates recognise square and triangular numbers and find the nth term of a linear sequence, but they are never asked to deduce a rule of the form $an^2 + bn + c$ .
Why do you halve the second difference?: Because the second difference of $an^2 + bn + c$ always works out as $2a$ . The $bn + c$ part has a constant first difference, so it adds nothing at the second stage, and $n^2$ on its own has second difference 2. A measured second difference of 6 therefore means $a = 3$ .
How many marks is a quadratic sequence question worth on Edexcel Higher?: Finding the nth term from a list of terms is usually 3 marks: one for the second difference and the $an^2$ it gives, one for the subtraction, one for the complete expression. Generating a term from a rule you are given is 1-2 marks, and the version that quotes the rule as $an^2 + bn$ and supplies two terms runs to 4 or 5.
Is the second difference always 2?: No, that only happens when the coefficient of $n^2$ is 1, as it is for the square numbers. A second difference of 8 gives $4n^2$ , a second difference of 1 gives $\frac{1}{2}n^2$ , which is how the triangular numbers 1, 3, 6, 10 behave, and a sequence that turns downwards has a negative second difference.

Quadratic Sequences - Edexcel GCSE Higher Maths

Worked example

Practice questions

Common mistakes examiners see

Frequently asked questions

Related topics