Mark scheme · Sequences & Patterns
Total: 20 marks
Award the marks shown for each correct step, then add up the total out of 20. A method mark counts even if the final answer is wrong.
A growing pattern uses counters.
Pattern 1 uses $6$ counters, Pattern 2 uses $11$ and Pattern 3 uses $16$.
How many more counters are needed to go from one pattern to the next?[2 marks]
MethodFind the difference between consecutive patterns: $11 - 6 = 5$.
Check: $16 - 11 = 5$. Each new pattern needs $5$ more.
Answer: $5$
Marks✔1 markDifference between consecutive patterns
✔1 markAnswer 5
A pattern is made from matchsticks.
Pattern 1 uses $6$ matchsticks, Pattern 2 uses $10$, Pattern 3 uses $14$ and Pattern 4 uses $18$.
How many matchsticks are used in Pattern 8?[3 marks]
MethodThe number of matchsticks goes up by $4$ from one pattern to the next.
Pattern 5 would use $22$, Pattern 6 $26$, and so on – each time add 4.
Continuing the pattern, Pattern 8 uses $34$ matchsticks.
Answer: $34$
Marks✔1 markCommon difference 4 between patterns
✔1 markContinue the count up to Pattern 8
✔1 markAnswer 34
Here is a sequence with one term missing.
$36,\ 31,\ \boxed{\phantom{00}},\ 21,\ 16$
Work out the value of the missing term.[2 marks]
MethodThe term-to-term rule is: subtract 5.
The missing term is $26$.
Answer: $26$
Marks✔1 markTerm-to-term rule: subtract 5
✔1 markMissing term 26
A growing pattern uses counters.
Pattern 1 uses $9$ counters, Pattern 2 uses $15$ and Pattern 3 uses $21$.
How many more counters are needed to go from one pattern to the next?[2 marks]
MethodFind the difference between consecutive patterns: $15 - 9 = 6$.
Check: $21 - 15 = 6$. Each new pattern needs $6$ more.
Answer: $6$
Marks✔1 markDifference between consecutive patterns
✔1 markAnswer 6
A pattern is made from matchsticks.
Pattern 1 uses $5$ matchsticks, Pattern 2 uses $8$, Pattern 3 uses $11$ and Pattern 4 uses $14$.
How many matchsticks are used in Pattern 8?[3 marks]
MethodThe number of matchsticks goes up by $3$ from one pattern to the next.
Pattern 5 would use $17$, Pattern 6 $20$, and so on – each time add 3.
Continuing the pattern, Pattern 8 uses $26$ matchsticks.
Answer: $26$
Marks✔1 markCommon difference 3 between patterns
✔1 markContinue the count up to Pattern 8
✔1 markAnswer 26
Here is a sequence with one term missing.
$2,\ 4,\ 8,\ \boxed{\phantom{00}},\ 32$
Work out the value of the missing term.[2 marks]
MethodThe term-to-term rule is: multiply by 2.
The missing term is $16$.
Answer: $16$
Marks✔1 markTerm-to-term rule: multiply by 2
✔1 markMissing term 16
A growing pattern uses square tiles.
Pattern 1 uses $6$ square tiles, Pattern 2 uses $10$ and Pattern 3 uses $14$.
How many more square tiles are needed to go from one pattern to the next?[2 marks]
MethodFind the difference between consecutive patterns: $10 - 6 = 4$.
Check: $14 - 10 = 4$. Each new pattern needs $4$ more.
Answer: $4$
Marks✔1 markDifference between consecutive patterns
✔1 markAnswer 4
A pattern is made from dots.
Pattern 1 uses $4$ dots, Pattern 2 uses $7$, Pattern 3 uses $10$ and Pattern 4 uses $13$.
How many dots are used in Pattern 8?[3 marks]
MethodThe number of dots goes up by $3$ from one pattern to the next.
Pattern 5 would use $16$, Pattern 6 $19$, and so on – each time add 3.
Continuing the pattern, Pattern 8 uses $25$ dots.
Answer: $25$
Marks✔1 markCommon difference 3 between patterns
✔1 markContinue the count up to Pattern 8
✔1 markAnswer 25
Here is a sequence with one term missing.
$7,\ \boxed{\phantom{00}},\ 19,\ 25,\ 31$
Work out the value of the missing term.[2 marks]
MethodThe term-to-term rule is: add 6.
The missing term is $13$.
Answer: $13$
Marks✔1 markTerm-to-term rule: add 6
✔1 markMissing term 13
A growing pattern uses matchsticks.
Pattern 1 uses $9$ matchsticks, Pattern 2 uses $15$ and Pattern 3 uses $21$.
How many more matchsticks are needed to go from one pattern to the next?[2 marks]
MethodFind the difference between consecutive patterns: $15 - 9 = 6$.
Check: $21 - 15 = 6$. Each new pattern needs $6$ more.
Answer: $6$
Marks✔1 markDifference between consecutive patterns
✔1 markAnswer 6
A pattern is made from counters.
Pattern 1 uses $6$ counters, Pattern 2 uses $11$, Pattern 3 uses $16$ and Pattern 4 uses $21$.
How many counters are used in Pattern 7?[3 marks]
MethodThe number of counters goes up by $5$ from one pattern to the next.
Pattern 5 would use $26$, Pattern 6 $31$, and so on – each time add 5.
Continuing the pattern, Pattern 7 uses $36$ counters.
Answer: $36$
Marks✔1 markCommon difference 5 between patterns
✔1 markContinue the count up to Pattern 7
✔1 markAnswer 36
In this number chain the same two steps repeat: “$\times 2$ then $-1$”, then “$+1$”, alternating.
The chain so far is:
$4, \; 7, \; 8, \; 15, \; 16, \; \boxed{?}$
The next step is “$\times 2$ then $-1$”.
Work out the next number in the chain.[3 marks]
MethodThe steps alternate: “$\times 2$ then $-1$”, then “$+1$”.
The next step is “$\times 2$ then $-1$”: $16 \times 2 - 1 = 31$.
Answer: $31$
Marks✔1 markIdentifies the next step is ×2 then −1
✔1 mark16 × 2 − 1
✔1 mark31
Each term of this Fibonacci-type sequence is the sum of the two terms before it.
$1,\ 5,\ \boxed{\phantom{0}},\ 11,\ 17$
Work out the missing term.[2 marks]
MethodEach term is the sum of the two before it, so a term equals the next term minus the term before it.
$11 - 5 = 6$.
Answer: $6$
Marks✔1 markUse term = next − previous (or build forward)
✔1 markMissing term 6
In this sequence, each term after the second is the sum of the two terms before it.
The 1st, 3rd, 4th and 5th terms are
$2, \; \boxed{?}, \; 9, \; 16, \; 25$
Work out the missing 2nd term.[3 marks]
MethodThe 3rd term $=$ 1st $+$ 2nd, so $9 = 2 + (\text{2nd term})$.
2nd term $= 9 - 2 = 7$.
Check: $7 + 9 = 16$ ✓.
Answer: $7$
Marks✔1 mark3rd = 1st + 2nd
✔1 mark2nd = 9 − 2
✔1 mark7
The $n$th term of a sequence is $n^2 + 2$.
Work out the smallest integer $n$ for which the $n$th term is greater than $500$.[2 marks]
MethodSolve $n^2 + 2 > 500$, i.e. $n^2 > 498$.
$\sqrt{498} \approx 22.3$, so the smallest whole number is $n = 23$.
Check: $23^2 + 2 = 531 > 500$, but $22^2 + 2 = 486 \le 500$.
Answer: $23$
Marks✔1 markn² > 498
✔1 markn = 23
In this number chain the same two steps repeat: “$\times 2$ then $-5$”, then “$+5$”, alternating.
The chain so far is:
$4, \; 3, \; 8, \; 11, \; 16, \; \boxed{?}$
The next step is “$\times 2$ then $-5$”.
Work out the next number in the chain.[3 marks]
MethodThe steps alternate: “$\times 2$ then $-5$”, then “$+5$”.
The next step is “$\times 2$ then $-5$”: $16 \times 2 - 5 = 27$.
Answer: $27$
Marks✔1 markIdentifies the next step is ×2 then −5
✔1 mark16 × 2 − 5
✔1 mark27
The $n$th term of a sequence is $n^2 + 2$.
Work out the smallest integer $n$ for which the $n$th term is greater than $600$.[2 marks]
MethodSolve $n^2 + 2 > 600$, i.e. $n^2 > 598$.
$\sqrt{598} \approx 24.5$, so the smallest whole number is $n = 25$.
Check: $25^2 + 2 = 627 > 600$, but $24^2 + 2 = 578 \le 600$.
Answer: $25$
Marks✔1 markn² > 598
✔1 markn = 25
In this number chain the same two steps repeat: “$\times 4$ then $-1$”, then “$+1$”, alternating.
The chain so far is:
$4, \; 15, \; 16, \; 63, \; 64, \; \boxed{?}$
The next step is “$\times 4$ then $-1$”.
Work out the next number in the chain.[3 marks]
MethodThe steps alternate: “$\times 4$ then $-1$”, then “$+1$”.
The next step is “$\times 4$ then $-1$”: $64 \times 4 - 1 = 255$.
Answer: $255$
Marks✔1 markIdentifies the next step is ×4 then −1
✔1 mark64 × 4 − 1
✔1 mark255
Each term of this Fibonacci-type sequence is the sum of the two terms before it.
$3,\ 5,\ 8,\ \boxed{\phantom{0}},\ 21$
Work out the missing term.[2 marks]
MethodEach term is the sum of the two before it, so a term equals the next term minus the term before it.
$21 - 8 = 13$.
Answer: $13$
Marks✔1 markUse term = next − previous (or build forward)
✔1 markMissing term 13
In this sequence, each term after the second is the sum of the two terms before it.
The 1st, 3rd, 4th and 5th terms are
$5, \; \boxed{?}, \; 14, \; 23, \; 37$
Work out the missing 2nd term.[3 marks]
MethodThe 3rd term $=$ 1st $+$ 2nd, so $14 = 5 + (\text{2nd term})$.
2nd term $= 14 - 5 = 9$.
Check: $9 + 14 = 23$ ✓.
Answer: $9$
Marks✔1 mark3rd = 1st + 2nd
✔1 mark2nd = 14 − 5
✔1 mark9
The $n$th term of a sequence is $n^2 + 2$.
Work out the smallest integer $n$ for which the $n$th term is greater than $300$.[2 marks]
MethodSolve $n^2 + 2 > 300$, i.e. $n^2 > 298$.
$\sqrt{298} \approx 17.3$, so the smallest whole number is $n = 18$.
Check: $18^2 + 2 = 326 > 300$, but $17^2 + 2 = 291 \le 300$.
Answer: $18$
Marks✔1 markn² > 298
✔1 markn = 18
In this number chain the same two steps repeat: “$\times 4$ then $-1$”, then “$+1$”, alternating.
The chain so far is:
$2, \; 7, \; 8, \; 31, \; 32, \; \boxed{?}$
The next step is “$\times 4$ then $-1$”.
Work out the next number in the chain.[3 marks]
MethodThe steps alternate: “$\times 4$ then $-1$”, then “$+1$”.
The next step is “$\times 4$ then $-1$”: $32 \times 4 - 1 = 127$.
Answer: $127$
Marks✔1 markIdentifies the next step is ×4 then −1
✔1 mark32 × 4 − 1
✔1 mark127
The population of a town is modelled by $P_{n+1} = a\,P_n + 800$, where $P_n$ is the population at the start of year $n$.
At the start of year 1 the population is $30000$, and at the start of year 2 it is $33800$.
Use the model to work out the population at the start of year 4.[3 marks]
MethodFind $a$: $33800 = a\times30000 + 800$, so $a = (33800 - 800) \div 30000 = 1.1$.
Year 3: $P_3 = 1.1\times33800 + 800 = 37980$.
Year 4: $P_4 = 1.1\times37980 + 800 = 42578$.
Answer: $42578$
Marks✔1 marka = 1.1
✔1 markyear 3 population 37980
✔1 markyear 4 population 42578
The population of a town is modelled by $P_{n+1} = a\,P_n + 500$, where $P_n$ is the population at the start of year $n$.
At the start of year 1 the population is $40000$, and at the start of year 2 it is $48500$.
Use the model to work out the population at the start of year 4.[3 marks]
MethodFind $a$: $48500 = a\times40000 + 500$, so $a = (48500 - 500) \div 40000 = 1.2$.
Year 3: $P_3 = 1.2\times48500 + 500 = 58700$.
Year 4: $P_4 = 1.2\times58700 + 500 = 70940$.
Answer: $70940$
Marks✔1 marka = 1.2
✔1 markyear 3 population 58700
✔1 markyear 4 population 70940
The population of a town is modelled by $P_{n+1} = a\,P_n + 1000$, where $P_n$ is the population at the start of year $n$.
At the start of year 1 the population is $20000$, and at the start of year 2 it is $25000$.
Use the model to work out the population at the start of year 4.[3 marks]
MethodFind $a$: $25000 = a\times20000 + 1000$, so $a = (25000 - 1000) \div 20000 = 1.2$.
Year 3: $P_3 = 1.2\times25000 + 1000 = 31000$.
Year 4: $P_4 = 1.2\times31000 + 1000 = 38200$.
Answer: $38200$
Marks✔1 marka = 1.2
✔1 markyear 3 population 31000
✔1 markyear 4 population 38200
The population of a town is modelled by $P_{n+1} = a\,P_n + 1000$, where $P_n$ is the population at the start of year $n$.
At the start of year 1 the population is $30000$, and at the start of year 2 it is $34000$.
Use the model to work out the population at the start of year 4.[3 marks]
MethodFind $a$: $34000 = a\times30000 + 1000$, so $a = (34000 - 1000) \div 30000 = 1.1$.
Year 3: $P_3 = 1.1\times34000 + 1000 = 38400$.
Year 4: $P_4 = 1.1\times38400 + 1000 = 43240$.
Answer: $43240$
Marks✔1 marka = 1.1
✔1 markyear 3 population 38400
✔1 markyear 4 population 43240
The population of a town is modelled by $P_{n+1} = a\,P_n + 1000$, where $P_n$ is the population at the start of year $n$.
At the start of year 1 the population is $25000$, and at the start of year 2 it is $31000$.
Use the model to work out the population at the start of year 4.[3 marks]
MethodFind $a$: $31000 = a\times25000 + 1000$, so $a = (31000 - 1000) \div 25000 = 1.2$.
Year 3: $P_3 = 1.2\times31000 + 1000 = 38200$.
Year 4: $P_4 = 1.2\times38200 + 1000 = 46840$.
Answer: $46840$
Marks✔1 marka = 1.2
✔1 markyear 3 population 38200
✔1 markyear 4 population 46840
The population of a town is modelled by $P_{n+1} = a\,P_n + 800$, where $P_n$ is the population at the start of year $n$.
At the start of year 1 the population is $40000$, and at the start of year 2 it is $48800$.
Use the model to work out the population at the start of year 4.[3 marks]
MethodFind $a$: $48800 = a\times40000 + 800$, so $a = (48800 - 800) \div 40000 = 1.2$.
Year 3: $P_3 = 1.2\times48800 + 800 = 59360$.
Year 4: $P_4 = 1.2\times59360 + 800 = 72032$.
Answer: $72032$
Marks✔1 marka = 1.2
✔1 markyear 3 population 59360
✔1 markyear 4 population 72032
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