Sequences: nth term (linear & quadratic), geometric, Fibonacci-type – GCSE Maths exam questions with mark scheme

1Grade 1–3FoundationNon-calculator4 marks

Here are the first four terms of an arithmetic sequence.

$$5 \qquad 9 \qquad 13 \qquad 17$$

(a) Write down the next term. [1]

(b) Find an expression, in terms of $n$, for the $n$th term of the sequence. [2]

(a) $21$B1
(b) for $4n + k$ where $k \ne 1$ (or for $4n$ seen)B1
(b) (total B2) for $4n + 1$ oeB1
(c) for a correct reason, e.g. 'all the terms are odd and 98 is even', or $4n + 1 = 98$ gives $n = 24.25$ which is not a whole number, or '97 and 101 are terms, so 98 is not'C1
Answer: (a) $21$ (b) $4n + 1$ (c) e.g. every term is odd but 98 is even
Examiner tip: In (c) 'it's not in the pattern' is not enough – give a mathematical reason, such as solving $4n + 1 = 98$.

2Grade 1–3FoundationNon-calculator2 marks

The first term of a sequence is 2.
The term-to-term rule is: multiply by 3 then subtract 1.

Work out the 4th term of the sequence. [2]

for applying the rule correctly at least once, e.g. 2nd term $= 5$ or 3rd term $= 14$M1
$41$A1
Answer: $41$
Examiner tip: The first term counts as term 1 – applying the rule four times gives the 5th term (122), not the 4th.

3Grade 4–5Non-calculator3 marks

Here are the first four terms of a geometric sequence.

$$3 \qquad 6 \qquad 12 \qquad 24$$

(a) Work out the 8th term of the sequence. [2]

(b) Which term of the sequence is equal to 1536? [1]

(a) for identifying the ratio 2 and continuing the sequence, e.g. 48, 96, … or $3 \times 2^7$M1
$384$A1
(b) $10$th (term)B1
Answer: (a) $384$ (b) the 10th term
Examiner tip: Check whether a sequence is arithmetic (add) or geometric (multiply) before you continue it – adding 6 gives the wrong 8th term.

4Grade 6–7HigherNon-calculator4 marks

The rule for a sequence is: add the two previous terms to get the next term.

The first term is $a$ and the second term is $b$.
The 3rd term is 11 and the 6th term is 47.

Find the values of $a$ and $b$. [4]

for writing the terms in terms of $a$ and $b$, at least as far as the 6th: $a,\ b,\ a + b,\ a + 2b,\ 2a + 3b,\ 3a + 5b$M1
for forming two correct equations: $a + b = 11$ and $3a + 5b = 47$M1
for a correct method to solve the simultaneous equations, e.g. $3a + 5(11 - a) = 47$ or $3a + 3b = 33$ subtractedM1
$a = 4$ and $b = 7$A1
Answer: $a = 4$, $b = 7$
Examiner tip: Write every term in terms of $a$ and $b$ carefully – one slip in the coefficients (e.g. $3a + 4b$) wrecks the equations. Check: 4, 7, 11, 18, 29, 47.

5Grade 6–7HigherNon-calculator3 marks

Here are the first five terms of a quadratic sequence.

$$4 \qquad 11 \qquad 22 \qquad 37 \qquad 56$$

Find an expression, in terms of $n$, for the $n$th term of this sequence. [3]

for finding the second difference of 4 and deducing $2n^2$M1
for subtracting $2n^2$ from the terms to get $2, 3, 4, 5, 6$ (or for a linear sequence that leads to $n + 1$)M1
$2n^2 + n + 1$A1
Answer: $2n^2 + n + 1$
Examiner tip: The coefficient of $n^2$ is HALF the second difference – a second difference of 4 gives $2n^2$, not $4n^2$.

6Grade 8–9HigherNon-calculator4 marks

The $n$th term of a sequence is $n^2 + 4n$

(a) Show that the difference between the $n$th term and the $(n + 1)$th term is $2n + 5$ [2]

(b) Two consecutive terms of the sequence have a difference of 47.
Work out the larger of these two terms. [2]

(a) for $(n + 1)^2 + 4(n + 1) - (n^2 + 4n)$ with $(n + 1)^2$ expanded correctlyM1
(a) for $n^2 + 2n + 1 + 4n + 4 - n^2 - 4n = 2n + 5$ shown with no errorsA1
(b) for solving $2n + 5 = 47$ to get $n = 21$M1
$572$A1
Answer: (a) $(n+1)^2 + 4(n+1) - (n^2 + 4n) = 2n + 5$ (shown) (b) $572$
Examiner tip: $n = 21$ gives the SMALLER term (525) – the larger one is the 22nd term: $22^2 + 4 \times 22 = 572$.

7Grade 6–7HigherNon-calculator3 marks

The first two terms of a geometric sequence are $\sqrt{2}$ and $2$.

(a) Work out the 7th term. Give your answer in the form $a\sqrt{2}$, where $a$ is an integer. [2]

(b) Which term of the sequence is equal to 64? [1]

(a) for finding the common ratio $\sqrt{2}$, e.g. $\frac{2}{\sqrt{2}}$, or listing terms $\sqrt{2},\ 2,\ 2\sqrt{2},\ 4,\ \ldots$M1
$8\sqrt{2}$A1
(b) the $12$th (term)B1
Answer: (a) $8\sqrt{2}$ (b) the 12th term
Examiner tip: Each term is $(\sqrt{2})^n$, so the even-numbered terms are whole powers of 2 – use that pattern rather than decimals.

8Grade 1–3FoundationNon-calculator3 marks

Here are the first three patterns in a sequence of patterns made from sticks.

Each pattern is a row of hexagons. Next to each other, hexagons share one stick.

(a) How many sticks are needed for Pattern 5? [1]

(b) Lena has 80 sticks.
Show that she cannot use all 80 sticks to make one of the patterns in this sequence. [2]

(a) $26$B1
(b) for a correct method, e.g. finding the rule $5n + 1$, or continuing the sequence 6, 11, 16, … at least as far as 76 and 81, or $(80 - 1) \div 5 = 15.8$ oeM1
(b) for a correct conclusion with full working, e.g. '76 and 81 are consecutive terms, so 80 is not a term', or '$5n + 1 = 80$ gives $n = 15.8$, which is not a whole number', or 'every term ends in 1 or 6, 80 does not'C1
Answer: (a) $26$ (b) e.g. $5n + 1 = 80$ gives $n = 15.8$, not a whole number (Pattern 15 uses 76, Pattern 16 uses 81)
Examiner tip: Count the sticks carefully – Pattern 1 has 6 but each extra hexagon only adds 5 because it shares a side.

9Grade 1–3FoundationNon-calculator5 marks

(a) Here are the first five triangular numbers.

$$1 \qquad 3 \qquad 6 \qquad 10 \qquad 15$$

Write down the next two triangular numbers. [1]

(b) Here is a list of numbers.

$$2 \qquad 12 \qquad 16 \qquad 21 \qquad 25 \qquad 27 \qquad 35$$

From the list, write down

(i) a cube number [1]

(ii) a square number. [1]

$$4 \qquad \ldots\ldots \qquad \ldots\ldots \qquad 18 \qquad 29$$

Fill in the two missing terms. [2]

(a) $21$ and $28$ (both needed)B1
(b)(i) $27$ caoB1
(b)(ii) $16$ or $25$ (either one; do not award if a non-square number is also given)B1
(c) for a correct method, e.g. a trial with a 2nd term that is checked through the rule, or $4 + 2x = 18$ where $x$ is the 2nd term, or working backwards $29 - 18 = 11$ then $18 - 11 = 7$M1
(c) $7$ and $11$ in the correct positionsA1
Answer: (a) $21,\ 28$ (b)(i) $27$ (ii) $16$ or $25$ (c) $4,\ 7,\ 11,\ 18,\ 29$
Examiner tip: In (c) work backwards from the end: the term before 29 is $29 - 18 = 11$, then the one before that is $18 - 11 = 7$.

10Grade 4–5Non-calculator4 marks

Here are the first four terms of an arithmetic sequence.

$$47 \qquad 41 \qquad 35 \qquad 29$$

(a) Find an expression, in terms of $n$, for the $n$th term of the sequence. [2]

(b) Find the first term of the sequence that is negative. [2]

(a) for $-6n + k$ where $k \ne 53$, or for $53 - kn$ where $k \ne 0$ (accept $-6n$ seen)B1
(a) (total B2) for $53 - 6n$ oe, e.g. $-6n + 53$B1
(b) for a correct method, e.g. $53 - 6n < 0$ or $53 - 6n = 0$ leading to $n = 8.8\ldots$, or listing terms continuing to at least 5 (the 8th term)M1
(b) $-1$ cao (ignore any reference to it being the 9th term; 'the 9th term' alone scores M1 only)A1
Answer: (a) $53 - 6n$ (b) $-1$
Examiner tip: Decreasing sequences have a negative number in front of $n$: write $53 - 6n$, not $6n + 53$ or $6n - 53$.

11Grade 4–5Non-calculator3 marks

The $n$th term of sequence A is $3n + 2$
The $n$th term of sequence B is $7n - 1$

Find all the numbers less than 60 that are in both sequences.
You must show your working. [3]

for generating terms of both sequences, at least 4 correct terms of each (A: 5, 8, 11, 14, …; B: 6, 13, 20, 27, …)P1
for identifying $20$ as a common termA1
for $20$ and $41$ only (no extra values), allow 20 and 41 found by noting common terms go up by 21A1
Answer: $20$ and $41$
Examiner tip: Write both lists out far enough – after the first match the common terms go up by $3 \times 7 = 21$, so don't stop at 20.

12Grade 4–5Calculator4 marks

Here are the first three terms of a geometric sequence.

$$256 \qquad 192 \qquad 144$$

(a) Write down the term-to-term rule. [1]

(b) Work out the 6th term of the sequence. [1]

(a) multiply by $0.75$ oe, e.g. $\times \frac{3}{4}$ (do not accept 'subtract 64' or 'divide by 1.33')B1
(b) $60.75$ oe, e.g. $60\frac{3}{4}$B1
(c) for continuing the sequence correctly to at least the 9th term ($25.6\ldots$), or for $256 \times 0.75^{n-1}$ evaluated for $n = 9$ or $n = 10$M1
(c) 10th term, with $19.2\ldots$ (or $19.22$) seenA1
Answer: (a) $\times 0.75$ (b) $60.75$ (c) the 10th term ($19.22\ldots$)
Examiner tip: Check the rule works for every pair: $256 - 64 = 192$ but $192 - 64 \ne 144$, so it must be a multiplier, not a difference.

13Grade 6–7HigherNon-calculator4 marks

Here are the first five terms of a quadratic sequence.

$$6 \qquad 9 \qquad 10 \qquad 9 \qquad 6$$

(a) Find an expression, in terms of $n$, for the $n$th term of this sequence. [3]

(b) Work out the 12th term of the sequence. [1]

(a) for second differences of $-2$ found (first differences 3, 1, −1, −3), or $-n^2$ seen as part of an expressionM1
(a) for subtracting $-n^2$ from the terms to get $7,\ 13,\ 19,\ldots$ (or an $an^2+bn+c$ method with a correct pair of simultaneous equations)M1
(a) $-n^2 + 6n + 1$ oeA1
(b) $-71$ ft their quadratic expression from (a), provided it is of the form $an^2 + bn + c$ with $a \ne 0$B1ft
Answer: (a) $-n^2 + 6n + 1$ (b) $-71$
Examiner tip: A negative second difference means a negative $n^2$ term – and when you subtract $-n^2$ you add $n^2$ to each term.

14Grade 6–7HigherNon-calculator3 marks

The 2nd term of a geometric sequence is 12
The 5th term of the same sequence is 40.5

All the terms of the sequence are positive.

Work out the 1st term of the sequence. [3]

for $12r^3 = 40.5$ or $r^3 = 40.5 \div 12$ oe (= 3.375 or $\frac{27}{8}$)M1
for $r = 1.5$ oeA1
$8$ cao (allow $12 \div 1.5$ seen with answer 8)A1
Answer: $8$ (common ratio $1.5$)
Examiner tip: From the 2nd to the 5th term you multiply by $r$ three times, not four – so it is $r^3$, not $r^4$.

15Grade 8–9HigherNon-calculator6 marks

The $n$th term of a sequence is $an^2 + bn$, where $a$ and $b$ are integers.

The 2nd term of the sequence is 2
The 5th term of the sequence is 35

(a) Find the value of $a$ and the value of $b$. [4]

(b) Is 189 a term in this sequence?
You must show how you get your answer. [2]

(a) for one correct equation, e.g. $4a + 2b = 2$ or $25a + 5b = 35$M1
(a) for both equations and a correct method to eliminate one unknown, e.g. $b = 1 - 2a$ substituted, or $10a + 5b = 5$ subtracted from $25a + 5b = 35$M1
(a) $a = 2$A1
(a) $b = -3$A1
(b) for $2n^2 - 3n = 189$ (ft their $a, b$) and a method to solve, e.g. $(2n - 21)(n + 9) = 0$, or for working out the 10th and 11th terms (170 and 209)M1
(b) No, with correct supporting values, e.g. $n = 10.5$ is not an integer, or '170 and 209 are consecutive terms'C1
Answer: (a) $a = 2,\ b = -3$ (b) No: $2n^2 - 3n = 189$ gives $n = 10.5$ (10th term is 170, 11th is 209)
Examiner tip: Substitute the position into the formula ($n = 2$ gives $4a + 2b$), not the term value, then solve the pair of equations.

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