Year 10 sequences worksheet answers to download as PDF

These Year 10 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.

← Back to the worksheet

Answer to Question 1
1. 21, 25. Add 4 each time. The difference between consecutive terms is always 4, so this is an arithmetic sequence: \(17+4=21\) and \(21+4=25\).
2. 5, 2.5. Divide by 2 (halve) each time. Each term is half of the one before, so the next terms are \(10\div2=5\) and \(5\div2=2.5\).
3. 162, 486. Multiply by 3 each time. Each term is 3 times the one before, so this is a geometric sequence: \(54\times3=162\).
4. 25, 36. These are the square numbers \(1^2,2^2,3^2,\ldots\) The differences are 3, 5, 7, so the next differences are 9 and 11: \(16+9=25\) and \(25+11=36\).
Answer to Question 2
1. Common difference 4, so \(4n+\ldots\): \(4\times1=4\), \(7-4=3\). \(4n+3\) The common difference tells us the number in front of \(n\). The zero term (the term before the first) is \(7-4=3\). Check: \(4\times2+3=11\).
2. Difference \(-3\): \(-3n+\ldots\), \(20-(-3)=23\). \(23-3n\) The terms go down by 3, so the coefficient of \(n\) is \(-3\). The zero term is \(20+3=23\). Check: \(23-3\times2=17\).
3. Difference 5: \(5n+\ldots\), \(-3-5=-8\). \(5n-8\) The terms go up by 5, and the zero term is \(-3-5=-8\). Check: \(5\times3-8=7\).
4. \(4\times50+3=\) 203 Substitute \(n=50\) into \(4n+3\). The 50th term is 203.
Answer to Question 3
1. \(n=1,2,3,4\): 3, 8, 13, 18 Substitute \(n=1\), 2, 3 and 4 into \(5n-2\); the terms go up by 5 each time.
2. 4, 7, 12, 19 Square \(n\) first, then add 3: \(1+3\), \(4+3\), \(9+3\), \(16+3\).
3. 2, 4, 8, 16 This is a geometric sequence where each term is double the one before.
4. 2, 6, 12, 20 Multiply each \(n\) by the next whole number: \(1\times2\), \(2\times3\), \(3\times4\), \(4\times5\).
Answer to Question 4
1. \(3n+1=100\), \(3n=99\), \(n=33\). Yes, the 33rd term. Set the \(n\)th term equal to the number and solve for \(n\). A number is a term only if \(n\) is a positive whole number.
2. \(4n-2=150\), \(4n=152\), \(n=38\). Yes, the 38th term.
3. \(6n+1=75\), \(6n=74\), \(n=12.33\), which is not a whole number. No.
4. \(4n+3=203\), \(4n=200\), \(n=50\). Yes, the 50th term.
Answer to Question 5
1. Ratio 2. 10th term \(=3\times2^9=3\times512=\) 1536 Each term is double the one before. The \(n\)th term of a geometric sequence is \(a\times r^{n-1}\), so the 10th term is \(3\times2^9\).
2. Ratio \(\tfrac12\). 6th term \(=160\times\left(\tfrac12\right)^5=\) 5 Each term is half the one before, so the ratio is \(\tfrac12\). The 6th term is \(160\times\left(\tfrac12\right)^5=160\div32\).
3. Terms: 81, 243, 729, 2187. The first term bigger than 1000 is 2187. Keep multiplying by 3: 729 is still less than 1000, so the first term bigger than 1000 is 2187.
4. Ratio \(-2\). Next terms: −16, 32. Each term is multiplied by \(-2\), so the signs alternate between positive and negative.
Answer to Question 6
1. Triangular numbers: 21, 28 The differences increase by 1 each time: \(15+6=21\) and \(21+7=28\).
2. Cube numbers: 125, 216 These are \(1^3\), \(2^3\), \(3^3\), \(4^3\), so the next terms are \(5^3=125\) and \(6^3=216\).
3. Fibonacci sequence (add the previous two terms): 13, 21 Each term is the sum of the two before it: \(5+8=13\) and \(8+13=21\).
4. Prime numbers: 13, 17 Each number has exactly two factors. After 11 the next primes are 13 and 17.
Answer to Question 7
1. Add 2 each time: pattern 4 has 9 and pattern 5 has 11 matches. Each new triangle needs 2 more matches, because it shares one side with the triangle before it.
2. Difference 2, so \(2n+1\): \(2n+1\). The common difference 2 gives \(2n\), and the zero term is \(3-2=1\). Check: \(2\times3+1=7\).
3. \(2\times25+1=\) 51 matches. Substitute \(n=25\) into the \(n\)th term. Pattern 25 needs 51 matches.
4. \(2n+1\le100\), so \(n\le49.5\): pattern 49 uses \(99\) matches, leaving 1 match. Pattern 50 would need 101 matches, which is too many. So the biggest pattern is pattern 49, with 1 match left over.
Answer to Question 8
1. £65, £80, £95 She adds £15 each week: \(50+15=65\), \(65+15=80\), \(80+15=95\).
2. \(50+15n\) (an arithmetic sequence with first term 65 and difference 15, written \(15n+50\)) The common difference is 15 and the zero term is the starting £50.
3. \(50+15\times20=\) £350 Substitute \(n=20\). Ellie has £350 after 20 weeks.
4. \(50+15n>500\), \(15n>450\), \(n>30\), so 31 weeks. After 30 weeks she has exactly £500, which is not more than £500. So she first has more than £500 after 31 weeks.
Answer to Question 9
1. \(16+4=\) 20
2. The difference is 4, so \(4n+\ldots\) and \(8-4=4\). The \(n\)th term is \(4n+4\).
3. \(4n+4=60\), \(4n=56\), \(n=14\). White slabs \(=14^2=\) 196. So the patio has 14 white slabs on each side and 196 white slabs.
4. Each side has \(n\) slabs (not counting corners): \(4\times n\). The four corner slabs add 4 more: \(4n+4\).
Answer to Question 10
1. 2 tables: 10. 3 tables: 14. Each extra table adds 4 seats: \(6+4=10\) and \(10+4=14\).
2. Difference 4, first term 6, so \(4n+2\). The zero term is \(6-4=2\). Check: \(4\times3+2=14\).
3. \(4n+2\ge90\), \(4n\ge88\), \(n\ge22\): 22 tables (exactly 90). Solve the inequality. Exactly 22 tables seat \(4\times22+2=90\) people.
4. \(4\times12+2=\) 50 seats. Substitute \(n=12\) into the \(n\)th term. Twelve tables seat 50 people.
Answer to Question 11
1. \(100\times2^n\) The number starts at 100 and is multiplied by 2 every hour, so after \(n\) hours it has been multiplied by \(2^n\).
2. \(100\times256=\) 25 600 Substitute \(n=8\): \(2^8=256\). There are 25 600 bacteria after 8 hours.
3. \(100\times2^n>10^6\), \(2^n>10\,000\). \(2^{13}=8192\) and \(2^{14}=16\,384\). So 14 hours. Divide both sides by 100, then try powers of 2. After 13 hours there are only 819 200 bacteria, so the number first passes 1 million after 14 hours.
4. It is geometric because each term is multiplied by the same number, 2. An arithmetic sequence would add the same number each time instead.
Answer to Question 12
1. £1050, £1102.50, £1157.63 Each year the amount is multiplied by 1.05: \(1000\times1.05=1050\), \(1050\times1.05=1102.50\), and \(1102.50\times1.05=1157.625\), which rounds to £1157.63.
2. 1.05 A 5% increase means each term is 105% of the term before, so the common ratio is 1.05.
3. \(1000\times1.05^5=\) £1276.28 Multiply £1000 by \(1.05\) five times, which is \(1.05^5=1.27628\ldots\). Sam has £1276.28 after 5 years.
4. \(1.05^8=1.477\), \(1.05^9=1.551\). So 9 years. Try powers of 1.05 until the amount first passes £1500. After 8 years Sam has about £1477, which is not enough, so it takes 9 years.
Answer to Question 13
1. 6 tins The row number tells us how many tins are in that row, so the 6th row has 6 tins.
2. \(1+2+3+4+5+6=\) 21 Add the tins in all six rows. This total is the 6th triangular number, 21.
3. \(\dfrac{10\times11}{2}=\) 55 Substitute \(n=10\) into the formula. There are 55 tins in 10 rows.
4. \(n=19\): \(\dfrac{19\times20}2=190\); \(n=20\): \(210>200\). So 19 rows with 10 tins left over. Try values of \(n\) in the formula: 19 rows use 190 tins and 20 rows would need 210. So the shop makes 19 full rows with \(200-190=10\) tins left over.
Answer to Question 14
1. \(12,14,16,18,20\): 20 seats. Add 2 seats for each row: the rows have 12, 14, 16, 18 and 20 seats. Row 5 has 20 seats.
2. Difference 2, \(12-2=10\): \(2n+10\). The common difference is 2, and the zero term is \(12-2=10\). Check: \(2\times5+10=20\).
3. \(2\times20+10=\) 50 seats. Substitute \(n=20\) into the \(n\)th term. The back row has 50 seats.
4. \(12+14+16+18+20=\) 80 seats. Add the seats in the first 5 rows. There are 80 seats in the first 5 rows.
Answer to Question 15
1. Differences 6, 10, 14; second difference 4, so \(2n^2\). Subtract: \(3-2=1\), \(9-8=1\), \(19-18=1\). \(2n^2+1\)
2. Differences 5, 7, 9; second difference 2, so \(n^2\). Subtract: \(1,3,5\) which is \(2n-1\). \(n^2+2n-1\)
3. \(2\times400+1=\) 801 Substitute \(n=20\) into \(2n^2+1\): \(20^2=400\).
4. \(n^2+2n-1=62\), \(n^2+2n-63=0\), \((n+9)(n-7)=0\), so \(n=7\): yes, the 7th term (\(49+14-1=62\)).
Answer to Question 16
1. \(4n+3=4n+2+1=2(2n+1)+1\), which is 2 times a whole number plus 1: odd.
2. \(1=1^2\), \(1+3=4=2^2\), \(1+3+5=9=3^2\), \(1+3+5+7=16=4^2\) ✓
3. \((n+1)^2-n^2=2n+1\), which is odd.
4. \(6n-1=6(n-1)+5\), so the remainder after dividing by 6 is 5. It leaves remainder 2 when divided by 3 (\(6n-1=3(2n-1)+2\)), so it is never divisible by 3.
Answer to Question 17
1. 3rd: \(a+b\). 4th: \(a+2b\). 5th: \(2a+3b\). Each new term is the sum of the two before it, so collect the \(a\) and \(b\) terms each time.
2. \(a+2b=17\) and \(2a+3b=27\). From the first, \(a=17-2b\). Then \(34-4b+3b=27\), so \(b=7\) and \(a=3\). Solve the simultaneous equations by substitution. Check: \(3+14=17\) and \(6+21=27\).
3. 3, 7, 10, 17, 27, 44 Start with \(a=3\) and \(b=7\) and keep adding the last two terms.
4. 7th: \(27+44=71\), 8th: \(44+71=\) 115. Continue the sequence two more steps. The 8th term is 115.
Answer to Question 18
1. \(ar=12\) and \(ar^4=96\). The \(n\)th term of a geometric sequence is \(ar^{n-1}\), so the 2nd term is \(ar\) and the 5th term is \(ar^4\).
2. \(\dfrac{ar^4}{ar}=r^3=\dfrac{96}{12}=8\), so \(r=\) 2. Dividing cancels \(a\), leaving \(r^3=8\). The cube root of 8 is 2.
3. \(a=12\div2=\) 6. Terms: 6, 12, 24, 48, 96. Divide the second term by the ratio to get the first term, then keep doubling.
4. \(6\times2^9=6\times512=\) 3072. The 10th term is \(ar^9\). The 10th term is 3072.
Answer to Question 19
1. There are 50 pairs each adding to 101: \(50\times101=\) 5050. Pairing the first and last numbers always gives 101, and 100 numbers make 50 pairs. The sum is 5050.
2. \(3\times20+2=\) 62 Substitute \(n=20\) into \(3n+2\). The 20th term is 62.
3. \(S_{20}=\dfrac{20}{2}(5+62)=10\times67=\) 670 Use the formula with first term 5 and last term 62. The sum of the first 20 terms is 670.
4. \(S_n=\dfrac n2(5+3n+2)=\dfrac n2(3n+7)\). \(S_{25}=\dfrac{25}{2}\times82=1025>1000\) and \(S_{24}=12\times79=948\). So 25 terms. Try values of \(n\) in the formula for \(S_n\): 24 terms give 948 and 25 terms give 1025. So 25 terms are needed.
Answer to Question 20
1. First differences 7, 11, 15; second differences 4, 4. Subtract each term from the next: \(12-5=7\), \(23-12=11\), \(38-23=15\). The second differences are constant, so the sequence is quadratic.
2. Second difference 4, so \(2n^2\). Remainders: \(5-2=3\), \(12-8=4\), \(23-18=5\), which is \(n+2\). So \(2n^2+n+2\). Half of the second difference gives the coefficient of \(n^2\). Then the remainders 3, 4, 5 form the linear part \(n+2\).
3. \(2\times100+10+2=\) 212 Substitute \(n=10\) into \(2n^2+n+2\). Pattern 10 has 212 dots.
4. \(n=22\): \(968+22+2=992\). \(n=23\): \(1058+23+2=1083\). So pattern 23. Try values of \(n\) close to \(\sqrt{500}\approx22\). Pattern 22 has 992 dots and pattern 23 has 1083, so the first pattern with more than 1000 dots is pattern 23.

How to use these Year 10 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 10 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 10 sequences worksheet answers show every step, so nothing is left unexplained.