Year 7 sequences worksheet answers to download as PDF

These Year 7 sequences worksheet answers give a full worked solution to all 20 questions of the sequences worksheet. Try each question yourself first, then check your method step by step.

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Answer to Question 1
First look at how each term changes: a constant difference means add or subtract, a constant multiplier means multiply, and other patterns may be special sequences such as square or Fibonacci numbers.
1. 24, 29, 34. Add 5 each time.
2. 26, 20, 14. Subtract 6 each time.
3. 162, 486, 1458. Multiply by 3 each time.
4. 25, 36, 49. The square numbers.
5. 13, 21, 34. Add the two previous terms (Fibonacci).
Answer to Question 2
1. The difference is 3, so start with \(3n\) (3, 6, 9, …). The terms are 2 more: \(3n+2\).
2. Difference 5: \(5n\) gives 5, 10, 15, … The terms are 3 less: \(5n-3\).
3. Difference \(-3\): \(-3n\) gives \(-3,-6,-9,\ldots\) The terms are 23 more: \(-3n+23\) (or \(23-3n\)).
4. Difference 4: \(4n\) gives 4, 8, 12, … The terms are 3 less: \(4n-3\).
5. Difference 2: \(2n\) gives 2, 4, 6, … The terms are 5 more: \(2n+5\).
Check each rule with \(n=1\): for example \(3\times1+2=5\), the first term of (a).
Answer to Question 3
Substitute the position number for n to find a term. To test whether a number is in the sequence, set \(4n+3\) equal to it and check whether n is a whole number.
1. 7, 11, 15, 19.
2. \(4\times20+3=\)83.
3. \(4\times100+3=\)403.
4. \(4n+3=83\), \(4n=80\), \(n=20\). Yes, it is the 20th term.
5. \(4n+3=50\), \(4n=47\), \(n=11.75\) is not a whole number. No.
Answer to Question 4
1. 3, 5, 9, 17, 33.
2. The differences are 2, 4, 8, 16, which are not constant, so it is not linear.
3. 2, 8, 14, 20.
4. \(6n-4=50\), \(6n=54\), \(n=\)9.
A sequence is linear only when the difference between consecutive terms stays the same; here the differences double each time.
Answer to Question 5
1. Square numbers are \(n\times n\): 1, 4, 9, 16, 25, 36.
2. Triangular numbers add 2, then 3, then 4, …: 1, 3, 6, 10, 15.
3. Cube numbers are \(n\times n\times n\): 1, 8, 27, 64, 125.
4. Each power of 2 doubles the previous one: 2, 4, 8, 16, 32, 64.
5. \(49=7^{2}\), \(64=8^{2}\) and \(100=10^{2}\) are square. 50 and 120 lie between squares, so the square numbers are 49, 64 and 100.
Answer to Question 6
1. Three steps to go from 3 to 18: difference \(15\div3=5\). 8, 13.
2. Difference 6: 8 and 26.
3. Four steps from 50 to 10: difference \(-40\div4=-10\). 40, 30, 20.
4. Four steps from 11 to 31: difference 5. First term \(11-10=\)1.
Answer to Question 7
1. The amount goes up by £3 each week: week 4 is £14 and week 5 is £17.
2. Difference 3, so start with \(3n\) (3, 6, 9, …). Each amount is 2 more: \(3n+2\) pounds.
3. Substitute \(n=12\): \(3\times12+2=\)£38.
4. \(3n+2>50\) gives \(3n>48\), so \(n>16\). Week 16 is exactly £50, so the first week with more than £50 is week 17 (£53).
Answer to Question 8
1. Each new house adds 4 matchsticks: \(13+4=\)17.
2. Difference 4, so start with \(4n\). For \(n=1\), \(4n=4\) but we need 5, so add 1: \(4n+1\).
3. Substitute \(n=30\): \(4\times30+1=\)121 matchsticks.
4. \(4n+1\le150\), so \(4n\le149\) and \(n\le37.25\). She can build 37 houses using \(4\times37+1=149\) matchsticks, with 1 left over.
Answer to Question 9
1. Add 2 for each row: 12, 14, 16, 18, 20. Row 5 has 20 seats.
2. Difference 2, so \(2n\); for \(n=1\) we need 12, so add 10: \(2n+10\).
3. \(2n+10=40\), so \(2n=30\) and \(n=\)15.
4. The back row is row 15 with 40 seats. The rows go from 12 to 40, with an average of \(\dfrac{12+40}{2}=26\) seats, so the total is \(15\times26=\)390 seats.
Answer to Question 10
1. The costs go 4, 6.50, 9, 11.50: each is £2.50 more than the last, which matches the rule.
2. Difference 2.5, so \(2.5n\); for \(n=1\) this gives 2.5 but the cost is 4, so add 1.5: \(C=2.5n+1.5\).
3. \(2.5\times10+1.5=25+1.5=\)£26.50.
4. \(2.5n+1.5=34\), so \(2.5n=32.5\) and \(n=32.5\div2.5=\)13 miles.
Answer to Question 11
1. Year 3: 48. Year 4: 96.
2. No, it is geometric. The differences (6, 12, 24, 48) are not constant.
3. Year 5: 192. Year 6: 384.
4. Year 7: 768. Year 8: 1 536. After 8 years.
Doubling sequences are geometric: they grow slowly at first and then very quickly, which is why the rabbits pass 1 000 only two years after reaching about 400.
Answer to Question 12
The total number of tins for n rows is a triangular number, found by adding \(1+2+3+\cdots+n\).
1. \(1+2+3+4+5=\)15.
2. 8.
3. 1, 3, 6, 10, 15 (the triangular numbers).
4. 13 rows use 91 tins and 14 rows use 105 tins, so 13 rows with 9 tins left over.
Answer to Question 13
1. Add 7 cm each week: 19, 26, 33 and 40 cm.
2. Start height 12 cm plus 7 cm for each of the n weeks: \(7n+12\).
3. \(7\times10+12=\)82 cm.
4. 2 m = 200 cm. \(7n+12>200\) gives \(7n>188\) and \(n>26.86\). The first whole week is week 27 (201 cm).
Answer to Question 14
1. 4, 5.5, 7, 8.5, 10 km.
2. \(1.5n+2.5\).
3. \(1.5n+2.5>10\), \(n>5\). Week 6 (11.5 km).
4. \(1.5\times12+2.5=20.5\), so she runs 20.5 km, which is more than 20 km. The coach is approximately right.
Remember that “more than 10 km” rules out week 5, where the distance is exactly 10 km.
Answer to Question 15
To match a linear sequence with its rule, look at the difference (the number in front of n) and check the first term. A sequence that multiplies by the same number is a power rule.
1. A: \(4n-1\). B: \(2n+1\). C: \(3n\). D: \(3^{n}\).
2. A: 39. B: 21. C: 30. D: 59 049.
3. D has the largest 10th term (\(3^{10}=59049\)).
4. Difference 6 and first term 3: \(6n-3\).
Answer to Question 16
1. Black tiles are square numbers: pattern 4 has \(4^{2}=\)16. White tiles go up by 4: \(16+4=\)20.
2. White tiles: 8, 12, 16, 20, difference 4, so \(4n\) plus 4: \(4n+4\).
3. Black tiles form an n by n square: \(n^{2}\).
4. \(4n+4=36\), so \(4n=32\) and \(n=8\). Pattern 8 has \(8^{2}=\)64 black tiles.
Answer to Question 17
1. 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144.
2. \(1+1+2+3+5=12\). The 7th term is 13 and \(13-1=12\). They are equal.
3. 2, 8, 34, 144 (every third term is even).
4. 2, 5, 7, 12, 19, 31, 50, 81.
Answer to Question 18
1. The difference is 3, so the rule starts with 3n. \(7-3=4\), so the nth term is 3n + 4.
2. \(3\times10-2=\)28. Beth subtracted first.
3. \(4n+5=31\), \(4n=26\), \(n=6.5\). Carl divided by 2 instead of 4. 31 is not in the sequence.
4. The sequence doubles each time, so the next term is 32.
Answer to Question 19
1. Ali: 20, 25, 30, 35, 40, 45. Bea: 5, 13, 21, 29, 37, 45.
2. Ali: \(20+5n\). Bea: \(5+8n\).
3. \(20+5n=5+8n\), \(15=3n\), \(n=\)5 weeks (both have £45).
4. After 10 weeks: Ali £70, Bea £85. Bea has £15 more.
Bea saves more each week, so although she starts with less, her total overtakes Ali’s after week 5.
Answer to Question 20
1. Differences: 3, 5, 7. They are not constant, so the sequence is not linear.
2. Second differences: 2, 2.
3. Constant second differences of 2 tell you the nth term contains \(n^{2}\) (the coefficient is half of 2).
4. \(n^{2}+2=51\), \(n^{2}=49\), \(n=7\). Pattern 7.
Check: \(7^{2}+2=49+2=51\).

How to use these Year 7 sequences worksheet answers

First, try every question on the worksheet without looking. Then compare your working with the solution line by line, not only the final answer. When your answer is different, find the first step that changes, because that is where the mistake is. After that, read the matching part of the sequences lesson again. Finally, check that the method is secure with the quiz or the test.

Also, the Year 7 sequences worksheet answers are available as a PDF, so you can mark printed work quickly. Moreover, all the pages of this chapter work together, so you can move between them at any time. For example, many pupils read the lesson on Monday, practise on Wednesday and test themselves on Friday. In short, these Year 7 sequences worksheet answers show every step, so nothing is left unexplained.