How to use this menu
Six original tasks: two for criterion A (a short check and an end-of-unit test), one investigation for B, one communication task for C, and two real-life tasks for D that also assess C. Set at least one task for each criterion across the unit; nobody needs all six. The timings spread assessment through the unit rather than leaving it all to the end.
This is a Year 11 Extended unit, so the tasks are pitched at the MYP Year 5 descriptors: unfamiliar problems, general rules, justification and judgement of reasonableness, not a restatement of the lessons.
Descriptors: Year 11 Extended is assessed with the MYP Year 5 descriptors, paraphrased below each task. Check them against your school's current guide.
The menu at a glance
| Task | Criteria | When | Time | Format |
|---|---|---|---|---|
| A1: Quick check: arithmetic series and sigma | A | After lesson 3 | 25 minutes | Individual, written, in class. 16 marks. |
| A2: End-of-unit test | A | End of unit, after lesson 7 | 50 minutes | Individual, written, in class, one lesson. 30 marks in three sections. |
| B1: Running totals of cubes | B | Mid-unit, after lesson 3 | 50 minutes | In class. Students may collect results in pairs, but each writes up their own findings. |
| C1: Explain it to someone who missed it | C | After lesson 5 | 35 minutes | Individual, written, in class or as homework. One page of words, working and a worked check. |
| D1: The loom fund | D and C | After lesson 6 | 50 minutes | Individual, in class, one lesson. A written solution with explanations and a short note to the group treasurer. |
| D2: The clinic dosage plan | D and C | End of unit, after lesson 7 | 50 minutes | Individual, in class, one lesson. A written solution with explanations and a short reply to the head nurse. |
D1 and D2 are alternatives. Both assess criteria D and C, in two different real-life contexts, a village savings group and a clinic dosage plan: one is set, not both.
The unit's inquiry
- Key concept. Relationships: how the terms of a sequence, the sums built from them and the quantities in a formula depend on one another.
- Related concepts. Generalization, Patterns, Models
- Global context. Globalization and sustainability: how people save, borrow and dose medicine over time, and when repeated change settles at a limit.
- Statement of inquiry. Adding a pattern of terms follows a rule, and the rule shows both a total after n steps and the ceiling that a shrinking pattern settles under.
MYP Year 5 descriptors, paraphrased
Each task's clarification, on its student sheet below, says what these levels look like in that task.
| Level | Criterion A: Knowing and understanding |
|---|---|
| 1–2 | Selects suitable mathematics for simple problems in familiar situations and applies it with some success. |
| 3–4 | Selects suitable mathematics for more complex problems in familiar situations and applies it successfully. |
| 5–6 | Selects suitable mathematics for challenging problems in familiar situations, applies it successfully and generally solves these problems correctly in a range of contexts. |
| 7–8 | Selects suitable mathematics for challenging problems in both familiar and unfamiliar situations, applies it successfully and generally solves these problems correctly in a range of contexts. |
| Level | Criterion B: Investigating patterns |
|---|---|
| 1–2 | With support, applies problem-solving techniques to find simple patterns and states predictions that fit them. |
| 3–4 | Applies problem-solving techniques to find simple patterns and suggests general rules that fit the findings. |
| 5–6 | Applies problem-solving techniques to find complex patterns, describes them as general rules that fit the findings and verifies that the rules hold. |
| 7–8 | Selects and applies problem-solving techniques to find complex patterns, describes them as general rules that fit correct findings, and proves, or verifies and justifies, those rules. |
| Level | Criterion C: Communicating |
|---|---|
| 1–2 | Uses limited mathematical language and limited forms of representation; the reasoning is hard to follow. |
| 3–4 | Uses some appropriate mathematical language and forms of representation; the reasoning can be followed and the work is adequately organized. |
| 5–6 | Usually uses appropriate language, moves between forms of representation with some success, and the reasoning is complete and coherent in a mostly logical structure. |
| 7–8 | Consistently uses appropriate language, moves effectively between forms of representation, and the reasoning is complete, coherent and concise in a logical structure. |
| Level | Criterion D: Applying mathematics in real-life contexts |
|---|---|
| 1–2 | Identifies some relevant elements of the real-life situation and selects, with some success, strategies to model it. |
| 3–4 | Identifies the relevant elements, selects adequate strategies to model the situation, applies them to reach a solution and discusses whether it makes sense. |
| 5–6 | Identifies the relevant elements, selects adequate modelling strategies and applies them to reach a valid solution, explains the degree of accuracy and explains whether the solution makes sense. |
| 7–8 | Identifies the relevant elements, selects appropriate modelling strategies and applies them successfully to reach a correct solution, justifies the degree of accuracy and justifies whether the solution makes sense. |
Task A1: Quick check: arithmetic series and sigma
Why set it. A short check of lessons 1 to 3 while there is still time to reteach: the arithmetic sum, sigma notation and the partial sums. Questions 5 and 6 are unfamiliar, so the top band is open to everyone.
When: After lesson 3. Time: 25 minutes. Format: Individual, written, in class. 16 marks.
The student sheet, with its task-specific clarification
When: After lesson 3. Time: 25 minutes. Calculator: no calculator. Criterion: A, knowing and understanding.
Answer every question you can and show your working: a correct method earns credit even if the final answer slips. Questions 5 and 6 are meant to make you think, so try something rather than leaving them blank.
Part 1. Adding an arithmetic sequence
- Find the sum of the first 20 terms of 7, 11, 15, 19, … (2 marks)
- Find the sum of the series 5 + 9 + 13 + … + 101. Say how many terms it has. (3 marks)
Part 2. Sigma notation
- Write out the first three terms and the last term of Σk=112 (3k − 2), then find its value. (3 marks)
Part 3. Running totals
- The sum of the first n terms of an arithmetic sequence is Sn = n2 + 8n. Find the first term, the common difference and the 10th term. (3 marks)
Part 4. Think it through
- The first three terms of an arithmetic sequence are x, 2x + 1 and 4x − 2. Find x, then find the sum of the first 10 terms. (2 marks)
- Find a formula, in terms of n, for the sum of the first n even numbers 2 + 4 + 6 + … + 2n. Use it to find 2 + 4 + 6 + … + 100. (3 marks)
How your work is judged
| Level | Criterion A in task A1: what it looks like |
|---|---|
| 1–2 | You use a formula or a method correctly in some of questions 1 to 3, even if the number of terms or the arithmetic slips. |
| 3–4 | You find the number of terms and the sums in questions 1 to 3 correctly and make a start on question 4. |
| 5–6 | You answer questions 1 to 4 correctly with working that shows your method, and make a good start on question 5 or 6. |
| 7–8 | You answer questions 1 to 4 correctly and solve question 5 or 6 with reasoning that shows your answer works. |
Markscheme
| Task A1 question | Answer and marking |
|---|---|
| 1 | 900. S = 20 ÷ 2 × (2 × 7 + 19 × 4) = 10 × 90. 1 mark for the method, 1 for 900. The 20th term is 83, so 20 × (7 + 83) ÷ 2 = 900 gives the same. Using 20 × 4 in place of 19 × 4 gives 940, the usual slip. |
| 2 | 25 terms; sum 1325. n = (101 − 5) ÷ 4 + 1 = 25, then 25 ÷ 2 × (5 + 101) = 25 × 53. 1 mark for 25, 2 for 1325 (1 for the method). 24 terms, from forgetting the +1, gives 1272 and loses the first two marks. |
| 3 | 1, 4, 7, …, 34; value 210. 1 mark for the terms. The 12th term is 3 × 12 − 2 = 34; 12 terms, so 12 ÷ 2 × (1 + 34) = 6 × 35. 1 mark for the 12 terms or the last term, 1 for 210. Splitting also works: 3 × 78 − 2 × 12 = 234 − 24. |
| 4 | First term 9, difference 2, 10th term 27. S1 = 9, S2 = 20 so the second term is 11, so d = 2. 10th term = S10 − S9 = 180 − 153 = 27, or 9 + 9 × 2. 1 mark each. The second difference of Sn is 2, which is d: an equivalent route. |
| 5 | x = 4; sum 265. The two gaps are equal: (2x + 1) − x = (4x − 2) − (2x + 1), so x + 1 = 2x − 3 and x = 4 (1 mark). The terms are 4, 9, 14 with d = 5, and S10 = 5 × (8 + 9 × 5) = 5 × 53 = 265 (1 mark). Check: the 10th term is 49 and 5 × (4 + 49) = 265. |
| 6 | n(n + 1); 2550. Pairing, or 2 × (1 + 2 + … + n) = 2 × n(n + 1) ÷ 2, or n ÷ 2 × (2 + 2n). 1 mark for a method, 1 for n(n + 1). For 2 + … + 100, n = 50 and 50 × 51 = 2550: 1 mark. A formula of n2 (the odd numbers' sum) can be tested: n = 2 gives 6, not 4. |
| Marks to a level | Total 16. 1 to 4 marks: 1–2. 5 to 8: 3–4. 9 to 13: 5–6. 14 to 16, with question 5 or 6 fully correct: 7–8. Use the marks as a guide, then check the student's profile against the clarification before settling the level. |
Exemplar answers
| Task A1 response | What it looks like |
|---|---|
| Level 3–4 profile | Question 1 right (2). Question 2 with 24 terms and 12 × 106 = 1272 (1). Question 3 right (3). Question 4: 9 and 11 found, nothing more (1). Nothing on 5 or 6. 7 marks, and the profile matches 3–4: a method used, one counting slip, nothing unfamiliar. |
| Level 7–8 answer to question 6 | “Pair the first and last: 2 + 2n, and there are n/2 pairs, so the sum is n/2 × (2 + 2n) = n(n + 1). Check n = 3: 2 + 4 + 6 = 12 = 3 × 4. For 2 to 100, n = 50, so the sum is 50 × 51 = 2550.” |
| Level 7–8 answer to question 5 | “In an arithmetic sequence the gaps are the same, so (2x + 1) − x = (4x − 2) − (2x + 1). That gives x + 1 = 2x − 3, so x = 4. The terms are 4, 9, 14, which go up by 5. S10 = 10/2 × (2 × 4 + 9 × 5) = 5 × 53 = 265. Check: the 10th term is 4 + 45 = 49, and 5 × (4 + 49) = 265.” |
Differentiation
A way in for anyone stuck
- Give the two forms of the sum formula on a reference card; recalling the formula is not the point of the check.
- For question 3, offer a three-column table of k, 3k − 2 and the running total before the sum is asked for.
- Suggest, for question 6, writing 2 + 4 + 6 + 8 and pairing the ends before generalizing. Reading is not what is being assessed.
Stretch
- Question 6 again for the first n multiples of 5. What is the sum of 5 + 10 + … + 100? (5n(n + 1) ÷ 2; 1050.)
- Find the sum of all the whole numbers from 1 to 200 that are not multiples of 3. (20 100 − 6633 = 13 467.)
Students learning in English as an additional language
| Word in task A1 | Plain meaning |
|---|---|
| common difference | the amount added each time to get the next term |
| term | one number in the sequence |
| sigma | the Greek letter Σ, meaning “add up all of these” |
| partial sum | the total of the first n terms only |
| generalize | find a rule that works for every value of n |
Sentence starters for task A1:
- The first term is … and the common difference is …, so …
- There are … terms because …
- I checked my formula with n = …, which gives …
Task A2: End-of-unit test
Why set it. The whole unit at three levels of demand. Section 3 is unfamiliar on purpose: it is where the evidence for 7–8 comes from.
When: End of unit, after lesson 7. Time: 50 minutes. Format: Individual, written, in class, one lesson. 30 marks in three sections.
The student sheet, with its task-specific clarification
When: End of unit, after lesson 7. Time: 50 minutes. Calculator: scientific calculator. Criterion: A, knowing and understanding.
Three sections, from familiar questions to ones you have not met in this form. Show your working everywhere: it earns credit even when an answer slips. A scientific calculator is allowed.
Section 1. Familiar questions
- Find the sum of the first 8 terms of the geometric sequence 3, 6, 12, 24, … (2 marks)
- Find the sum to infinity of 18 + 6 + 2 + … (2 marks)
- Write the recurring decimal 0.727272… as a fraction in its simplest form, using a geometric series. (3 marks)
- Make l the subject of S = n/2 (a + l), where l is the last term. (2 marks)
- y varies inversely as the square of x. When x = 2, y = 12. Find y when x = 6. (2 marks)
Section 2. More complex questions
- A geometric series has first term 5 and common ratio 1.2. Find the smallest number of terms for which the sum is more than 100. Show the sums that decide it. (3 marks)
- An arithmetic sequence has third term 10 and ninth term 34. Find the sum of its first 20 terms. (3 marks)
- Show that Σk=1n (4k + 1) = 2n2 + 3n. Hence find the value of n for which the sum is 324. (4 marks)
Section 3. Unfamiliar questions
- The sum to infinity of a geometric series is 40 and the sum of its first two terms is 30. Find the first term and the common ratio, and check that your answer works. (3 marks)
- Eli says: “If you double the number of terms of an arithmetic series, the sum doubles.” Is Eli right? Give a test with numbers, and say exactly when his claim holds. (3 marks)
- The power P of a wind turbine varies as the cube of the wind speed v and as the square of the blade length b. The wind speed rises by 20% and the blades are made 10% shorter. By what factor does P change? Say whether it has gone up or down, and by about how much. (3 marks)
How your work is judged
| Level | Criterion A in task A2: what it looks like |
|---|---|
| 1–2 | You answer some Section 1 questions correctly, for example a sum or a rearrangement. |
| 3–4 | You answer most of Section 1 correctly and make progress in Section 2. |
| 5–6 | You answer Section 1 and most of Section 2 correctly, with clear methods, and make a start in Section 3. |
| 7–8 | You answer Sections 1 and 2 correctly and solve most of Section 3, with reasoning that shows why your answers hold. |
Markscheme
| Task A2 question | Answer and marking |
|---|---|
| 1 | 765. 3(28 − 1) ÷ (2 − 1) = 3 × 255. 1 mark for the method, 1 for 765. 768 is 3 × 28, the next term after the sum, not the sum. |
| 2 | 27. r = 1/3, 18 ÷ (1 − 1/3) = 18 ÷ 2/3. 1 mark for r, 1 for 27. −9 comes from the ratio upside down, 18 ÷ (1 − 3); 18 ÷ (1 + 1/3) = 13.5 uses 1 + r. |
| 3 | 8/11. 0.72 + 0.0072 + …: a = 0.72, r = 0.01 (1), sum = 0.72 ÷ 0.99 (1), = 72/99 = 8/11 (1). 72/99 unsimplified scores 2. |
| 4 | l = 2S/n − a, equivalently (2S − an) ÷ n. 1 mark for 2S = n(a + l) or 2S/n = a + l, 1 for the last step. Test: a = 3, n = 10, S = 210 gives l = 42 − 3 = 39, the 10th term of 3, 7, 11, …. 2S − a, from dividing only one side by n, is the usual slip. |
| 5 | y = 4/3. k = yx2 = 12 × 4 = 48, so y = 48 ÷ 36. 1 mark for k, 1 for 4/3. Or: x is 3 times as big, so y is divided by 9: 12 ÷ 9. 4 comes from dividing by 3. |
| 6 | 9 terms. 5(1.2n − 1) ÷ 0.2 = 25(1.2n − 1) > 100, so 1.2n > 5. S8 = 82.5 (1.28 = 4.30) and S9 = 104.0 (1.29 = 5.16). 1 mark for the inequality, 1 for the sums, 1 for 9. Trial of terms with the right method is full marks. |
| 7 | 800. a + 2d = 10 and a + 8d = 34, so 6d = 24, d = 4, a = 2 (2 marks, 1 for d). S20 = 10 × (4 + 76) = 800 (1 mark). The 20th term is 78 and 20 ÷ 2 × (2 + 78) = 800 also. |
| 8 | n = 12. Σ(4k + 1) = 4 × n(n + 1)/2 + n = 2n2 + 3n (2 marks: 1 for splitting, 1 for the result). 2n2 + 3n − 324 = 0 factorizes as (2n + 27)(n − 12) = 0 (1 mark). n = 12, and −13.5 is rejected (1 mark). |
| 9 | a = 20, r = 1/2, or a = 60, r = −1/2. a = 40(1 − r) and a(1 + r) = 30 give 40(1 − r)(1 + r) = 30, so 1 − r2 = 3/4 (1 mark). r = 1/2 gives a = 20 (1 mark): 20 + 10 = 30 and 20 ÷ 0.5 = 40. r = −1/2 gives a = 60 (1 mark): 60 − 30 = 30 and 60 ÷ 1.5 = 40. Both ratios are between −1 and 1, so both are valid; the third mark needs the negative root kept and checked. |
| 10 | Eli is right only when the common difference is 0. Test: 1, 2, 3, …: S5 = 15 and S10 = 55, not 30 (1 mark for a valid test, 1 for the numbers). In general S2n − 2Sn = n2d, here 25 = 52 × 1, which is 0 only when d = 0 (1 mark for the condition, with or without the algebra). |
| 11 | P is multiplied by 1.23 × 0.92 = 1.39968, up by about 40%. 1 mark for 1.23 = 1.728, 1 for 0.92 = 0.81, 1 for the product and the statement. 1.2 × 0.9 = 1.08 ignores the powers; 1.2 × 3 × 0.9 × 2 reads the powers as multipliers. |
| Marks to a level | Total 30 (11 + 10 + 9). 1 to 7 marks: 1–2. 8 to 15: 3–4. 16 to 23: 5–6. 24 to 30, with at least 5 of Section 3's 9 marks: 7–8. Check the profile against the clarification before settling the level. |
Exemplar answers
| Task A2 response | What it looks like |
|---|---|
| Level 5–6 profile | Section 1: 765 (2), 27 (2), 72/99 (2), question 4 stopped at 2S = n(a + l) (1), 4/3 (2): 9 marks. Section 2: 9 terms with no check (2), 800 (3), the sum formula but a slip in factorizing (2): 7 marks. Section 3: question 9 finds a = 20, r = 1/2 but drops r = −1/2 (2), question 11 right (3), question 10 not attempted (0). 21 marks, below the 24 needed for 7–8 even though 5 of Section 3’s 9 marks are in place, so 5–6: Sections 1 and 2 mostly secure, a start on Section 3. |
| Level 7–8 answer to question 10 | “Not always. Try 1, 2, 3, …: the sum of 5 terms is 15 and of 10 terms is 55, but 2 × 15 = 30. The extra 25 is n2d = 25 × 1. So S2n = 2Sn + n2d, and the sum doubles only if d = 0, when every term is the same.” |
| Level 7–8 answer to question 11 | “P ∝ v3b2. v is 1.2 times as big, so v3 is 1.728 times as big. b is 0.9 times as big, so b2 is 0.81 times as big. P is multiplied by 1.728 × 0.81 = 1.39968, about 40% more.” |
Differentiation
A way in for anyone stuck
- Read Section 3 aloud on request: reading is not what is being assessed.
- Offer the formulae sheet from lessons 1 to 7 for question 4 and question 6, since the rearranging, not the recall, is the target.
- For question 11, give a table with columns for v, b and P, with the multiplier of each, to fill in.
Stretch
- A geometric series has sum to infinity 40 and first term 10. Find the sum of the first n terms in terms of n, and the smallest n for which it is more than 39. (40(1 − 0.75n); 0.75n < 0.025 first at n = 13.)
- An arithmetic series begins with 5 and its first 20 terms have sum 450. Find the common difference. Is it a whole number? (d = 35/19, about 1.84: no.)
Students learning in English as an additional language
| Word in task A2 | Plain meaning |
|---|---|
| common ratio | the number you multiply by each time in a geometric sequence |
| sum to infinity | the number the totals get closer and closer to when you add for ever |
| subject | the letter on its own on one side of a formula |
| varies inversely | when one quantity gets bigger, the other gets smaller in the same proportion |
| factor | the number everything is multiplied by |
Sentence starters for task A2:
- The ratio is …, so the sum to infinity is …
- The common difference is … because …
- Eli is wrong for … because …, and right when …
Task B1: Running totals of cubes
Why set it. A pattern hunt that goes beyond the unit's sums. The sheet never names the pattern: students organize their results, find a rule, test it on a total they have not worked out, and then justify it.
When: Mid-unit, after lesson 3. Time: 50 minutes. Format: In class. Students may collect results in pairs, but each writes up their own findings.
The student sheet, with its task-specific clarification
In MYP Year 5, prompts you give (a table outline, a first question) are allowed at 1–2 and 3–4: note it on the student's work. Do not prompt towards the rule itself.
When: Mid-unit, after lesson 3. Time: 50 minutes. Calculator: scientific calculator. Criterion: B, investigating patterns.
Anyone can write down 13 + 23 + 33 + … for a few terms, but nobody wants to add 20 cubes one by one. This task is about finding a short way, and about knowing how far you can trust it.
Part 1. Collect results
- Make a table with a row for each n from 1 to 8. Record n, the cube n3 and the running total 13 + 23 + … + n3. Design the table yourself.
Part 2. Describe what you find
- Look at your running totals. Describe what you notice, in words, as a rule a classmate could use to find the running total for any n.
- Use your rule to predict the running totals for n = 9 and n = 10. Then add 93 and 103 to your table to check.
Part 3. Test your rule
- Without adding every cube, find the running total for n = 20. Show how your rule gives it.
- Choose a value of n that you have not used. Work the total out the long way and test your rule against it. Say how confident you now are, and why.
Part 4. Explain
- Explain why your rule works, or show that if your rule is right for one value of n then it must be right for the next.
- Rosa says: “Your rule is certain to give 13 + 23 + … for 100 terms, because it worked eight times.” Comment on Rosa’s reasoning.
How your work is judged
| Level | Criterion B in task B1: what it looks like |
|---|---|
| 1–2 | With help, you work out some cubes and running totals and say one thing you notice about the totals. |
| 3–4 | You complete your table, describe a rule that fits your totals, and use it to predict and check n = 9 and n = 10. |
| 5–6 | You organize your results clearly, describe a rule that fits every total, use it for n = 20, and test it on a total you did not use to find it, saying how far you trust it. |
| 7–8 | You describe a general rule, use it for large n, test it on new cases and justify it, for example by showing how the step from one total to the next fits it; you say what your evidence does and does not prove. |
Markscheme
| Task B1 question | Answer and marking |
|---|---|
| The pattern (do not share it before the task) | The running total of the cubes is the square of the running total of the whole numbers: (n(n + 1) ÷ 2)2. The totals 1, 9, 36, 100, 225, 441, 784, 1296 are 12, 32, 62, 102, 152, 212, 282, 362, where 1, 3, 6, 10, … are the running totals of 1, 2, 3, 4, … from lesson 1. |
| 1 | Cubes 1, 8, 27, 64, 125, 216, 343, 512. Running totals 1, 9, 36, 100, 225, 441, 784, 1296. Credit a sensible table; one slip that is then carried forward correctly is 1 mark off, not more. |
| 2 | Any correct description of a square-number pattern, for example “every total is a square number, and the number being squared goes 1, 3, 6, 10, 15, …, each one adding one more than the step before”. A rule that notices only that the totals are squares (without saying which square) fits the data but cannot yet predict: evidence for 3–4. |
| 3 | n = 9: 2 025 (452); n = 10: 3 025 (552). Check: 1296 + 729 = 2 025 and 2 025 + 1000 = 3 025. |
| 4 | 44 100. 1 + 2 + … + 20 = 210 (lesson 1), and 2102 = 44 100. A student who writes 20 × 21 ÷ 2 = 210 first shows the use of the unit's formula. |
| 5 | Any new n, with an honest comparison. For n = 12: total of cubes by hand 6084 = 782. Look for a calculation done the long way, a match stated, and a measured confidence: the rule fits every n tried, which is evidence, not proof. |
| 6 | At 7–8: the step from n to n + 1 adds (n + 1)3, and ((n + 1)(n + 2) ÷ 2)2 − (n(n + 1) ÷ 2)2 = (n + 1)2(4n + 4) ÷ 4 = (n + 1)3, so the rule continues for ever once it holds at n = 1. Accept a clear diagram argument or any correct algebra. Rosa’s claim: eight matches are evidence, not a proof; a pattern can hold for many cases and then fail, so a reason is needed. |
Exemplar answers
| Task B1 response | What it looks like |
|---|---|
| Level 3–4 | “Totals: 1, 9, 36, 100, 225, 441, 784, 1296. They are all square numbers: 1, 3 squared, 6 squared, 10 squared. The numbers being squared go 1, 3, 6, 10, so next is 15, and 15 squared is 225. For n = 9 the number being squared is 45, so the total is 2 025.” Table and rule both right and used for n = 9 and 10, but it is not tied to n and nothing is tested beyond the first predictions. |
| Level 7–8 | “The total for n is the square of 1 + 2 + … + n, which is n(n + 1)/2. For n = 20 that is 210, so the total is 44 100. I tested n = 12 the long way: 1 + 8 + 27 + 64 + 125 + 216 + 343 + 512 + 729 + 1000 + 1331 + 1728 = 6084 and 78 squared is 6084. It works for everything I tried, but that is evidence rather than proof. If the rule holds for n, going to n + 1 adds (n + 1) cubed, and the difference between the two squares works out as exactly that, so it holds for ever.” |
Differentiation
A way in for anyone stuck
- Give a three-column table outline for Part 1 and let the student write the headings.
- Suggest getting each running total from the one before by adding the next cube, rather than starting again.
- Prompt with a question, never the rule: “Which of your totals are numbers you already know a name for?”
Stretch
- Write each cube as a sum of odd numbers: 1, 3 + 5, 7 + 9 + 11, …. What is the first number in the cube of 6, and why? (31; the first odd number is n2 − n + 1.)
- Find a formula for the running total of the squares of the whole numbers, 1 + 4 + 9 + …. (n(n + 1)(2n + 1) ÷ 6; check n = 6: 91.)
Students learning in English as an additional language
| Word in task B1 | Plain meaning |
|---|---|
| cube | a number multiplied by itself twice: 43 = 4 × 4 × 4 |
| running total | the sum so far, adding one more term each time |
| predict | say what will happen before you check |
| rule | a sentence or formula that works every time |
| evidence | results that support a claim, but may not prove it |
Sentence starters for task B1:
- The running totals are … and I notice …
- My rule is: …. It gives … for n = …
- I am / am not sure the rule always works because …
Task C1: Explain it to someone who missed it
Why set it. Criterion C on its own: notation, structure and a line of reasoning a classmate can follow, built around two real counting errors from lessons 1 to 4.
When: After lesson 5. Time: 35 minutes. Format: Individual, written, in class or as homework. One page of words, working and a worked check.
The student sheet, with its task-specific clarification
When: After lesson 5. Time: 35 minutes. Calculator: no calculator. Criterion: C, communicating.
Nia was away for the lessons on arithmetic and geometric series. She tried two questions from a classmate’s notes and her working has mistakes in it.
Nia’s working
(a) Find the sum of 5 + 8 + 11 + … + 62.
Number of terms = (62 − 5) ÷ 3 = 19. Sum = 19/2 × (5 + 62) = 636.5.
(b) Find Σk=16 3 × 2k.
First term 3, ratio 2, so the sum = 3(26 − 1) ÷ (2 − 1) = 189.
Write a one-page explanation for Nia. She should be able to use it on her own, without asking anyone.
Part 1. Put it right
- Find every mistake in Nia’s working. For each one, say what she did, why it is wrong, and write the correct working.
Part 2. Teach the idea
- Explain to Nia how to count the terms of an arithmetic series from its first term, last term and common difference, and why your method works. Use a small example of your own.
- Explain how to read the first term and the number of terms from a sum written with Σ, using Nia’s question (b).
Part 3. Check your explanation
- Give Nia a quick test she can use on any answer she finds, to see whether it is sensible. Show it on both of her results.
- Read your page as if you were Nia. Check that each step is on its own line, that every symbol is introduced, and that the equals sign is only between equal things.
How your work is judged
| Level | Criterion C in task C1: what it looks like |
|---|---|
| 1–2 | Parts of your explanation can be followed, but steps or symbols are missing and the order is hard to follow. |
| 3–4 | You explain what went wrong in both questions with some correct notation and words; your steps are complete and in a sensible order, so Nia could follow most of it. |
| 5–6 | You correct both questions and explain why each was wrong; your steps are laid out one per line, you usually use correct notation and terms, and your example shows the idea. |
| 7–8 | Every step is explained and ordered so Nia could learn from it alone; your notation (sigma limits, equals signs, powers) and key words are correct throughout, and your quick test is clear and shown on both results. |
Markscheme
| Task C1 question | Answer and marking |
|---|---|
| The mistake in (a) | Number of terms = (last − first) ÷ d + 1 = 19 + 1 = 20: 19 is the number of steps between the terms, one fewer than the terms. Correct: 20 ÷ 2 × (5 + 62) = 10 × 67 = 670. Sanity: every term is a whole number, so the sum cannot be 636.5. |
| The mistake in (b) | The sum starts at k = 1, so the first term is 3 × 21 = 6, not 3 (3 would be k = 0). Correct: 6(26 − 1) ÷ (2 − 1) = 6 × 63 = 378: 6 + 12 + 24 + 48 + 96 + 192. Her 189 is the sum of the six terms from k = 0 to k = 5 (3 + 6 + … + 96): starting one term too early also stopped her one term too early. Sanity: the last term alone is 192, which is bigger than 189. |
| Teaching the count (Part 2) | Use a small case such as 5, 8, 11: last − first = 6, steps of 3 give 2 gaps, and 2 gaps separate 3 terms. A fence picture (posts and gaps) is a good model. Check that the example works when computed. |
| Quick test (Part 3) | Any valid test: terms are whole numbers so the sum is whole; the sum must be bigger than its largest term when all terms are positive; the sum of an arithmetic series is roughly the number of terms times the middle term (20 × 33.5 = 670). Accept a test shown on both results. |
| How to level | Level the page as a whole against the clarification, not by counting corrections. Notation and organization decide between 5–6 and 7–8 more often than the corrections do. Mathematical slips count only where they break the line of reasoning Nia has to follow; accuracy itself is criterion A. |
Exemplar answers
| Task C1 response | What it looks like |
|---|---|
| Level 3–4 extract | “In (a) she forgot the one: there are 20 terms not 19. The sum is 10 × 67 = 670. In (b) she should have used 6 as the first term because k starts at 1 so it is 3 × 2 = 6. Then it is 6 × 63 = 378.” Both corrections are right and the reasons are given, but the steps run on in a line, the rule for counting is not explained with an example and there is no check. |
| Level 7–8 extract | “Step 1. Terms go up by 3, from 5 to 62. 62 − 5 = 57 is the distance covered, and 57 ÷ 3 = 19 is the number of steps. Step 2. Steps are gaps between terms: 5, 8, 11 has 2 gaps and 3 terms, so the number of terms is the number of steps plus 1, which is 20. Step 3. S = 20/2 × (5 + 62) = 670. Check: 20 terms with a middle term of about 33.5 gives about 670; 636.5 cannot be right as all the terms are whole numbers.” Each step on its own line, the reason for the +1 stated, equals signs only between equal values. |
Differentiation
A way in for anyone stuck
- Give the counting rule from lesson 1 and the first-term rule for sigma as a reference card.
- Let students explain aloud to a partner first, then write down what they said.
- Offer the frame: “Nia wrote …. This is wrong because …. The correct working is ….”
Stretch
- Explain why the first term of Σk=06 3 × 2k is 3 and how many terms it has. (3, with 7 terms: 3 × 127 = 381.)
- Write a second version for a student one year younger who has not met sigma. What did you change, and why?
Students learning in English as an additional language
| Word in task C1 | Plain meaning |
|---|---|
| number of steps | how many times the common difference is added to get from the first term to the last |
| first term | the term where the sum starts |
| sensible | makes sense when you check it against the problem |
| explain | say what you did and why |
Sentence starters for task C1:
- Nia wrote …, but …
- First you …, because …
- A quick check is to see whether …
Task D1: The loom fund
Why set it. Geometric series, a savings model and its limits inside one real decision. An alternative to D2: set one of them.
When: After lesson 6. Time: 50 minutes. Format: Individual, in class, one lesson. A written solution with explanations and a short note to the group treasurer.
The student sheet, with its task-specific clarification
When: After lesson 6. Time: 50 minutes. Calculator: scientific calculator. Criteria: D, applying mathematics in real-life contexts; C, communicating.
A village weaving group wants to save 50 000 coins for a new loom. Each member’s share is paid into the group’s account as 2 000 coins at the end of every month. The account adds 0.5% interest at the end of every month, on everything in it.
Part 1. Building a model
- Work out the total in the account at the end of 3 months. Show how many months of interest each payment has earned.
- Show that the total after n months can be written as 400 000(1.005n − 1), explaining each step.
Part 2. The target
- Find the smallest whole number of months after which the account holds at least 50 000 coins. Show the two totals that decide it.
- Without any interest, how many months would it take? How much of the 50 000 coins is interest in your answer to question 3?
Part 3. What if things change?
- The bank cuts the interest to 0.3% a month. How many months does it take now? Is it better than no interest at all?
- A member says: “Saving for 24 months gives exactly twice as much as saving for 12.” Use your model to test the claim.
Part 4. Does it make sense?
- State what you assumed to make the model work, and say which assumption is most likely to be wrong in real life.
- Write a short note to the group treasurer with your recommendation, so that someone who has not seen your working can understand it.
How your work is judged
| Level | Criterion D in task D1: what it looks like |
|---|---|
| 1–2 | You work out some of the right numbers, for example the total after 3 months, but without a model or a decision. |
| 3–4 | You find the total after 3 months, a formula, and the number of months for the target, and say whether it makes sense. |
| 5–6 | You build and use the model correctly, test the claim and the lower rate with it, state your assumptions and say how they affect your answer. |
| 7–8 | You choose and use suitable methods throughout, reach correct answers, explain why the model and the target are reasonable and what could make them wrong, and justify your recommendation. |
| Level | Criterion C in task D1: what it looks like |
|---|---|
| 1–2 | Some working is shown, but it is hard to see how you reached the number of months. |
| 3–4 | Your working can be followed, with some correct notation and units in coins. |
| 5–6 | Your working is clear, usually in order with units and notation, and your note to the treasurer is mostly well organized. |
| 7–8 | Your working is clear, complete and well ordered, units and notation are always correct, and your note to the treasurer makes sense on its own. |
Markscheme
| Task D1 question | Answer and marking |
|---|---|
| 1 | 6030.05 coins. The first payment earns 2 months of interest, the second 1, the third 0: 2 000(1.0052 + 1.005 + 1) = 2 000 × 3.015025. A calculation giving 6000 ignores the interest, and 6060.20 treats the payments as made at the start of the months. |
| 2 | Geometric series with a = 2 000, r = 1.005. Total = 2 000(1.005n − 1) ÷ (1.005 − 1), and 2 000 ÷ 0.005 = 400 000. Credit the reversed order: the last payment earns nothing, the first earns n − 1 months. |
| 3 | 24 months. 400 000(1.00524 − 1) = 50 864 coins (50 863.91) and at 23 months 48 621 coins (48 620.81). Solving 1.005n > 1.125 gives n > 23.6, so 24. 23 is wrong because 48 621 is below the target. |
| 4 | 25 months without interest (2 000 × 25 = 50 000). Interest in 24 months: 50 864 − 48 000 = 2864 coins. Interest saves one month. |
| 5 | 25 months. At 0.3% the total after 24 months is 49 693 coins (just short) and after 25 months 51 842 coins. It takes the same 25 months as no interest at all, though the final total is 1842 coins higher than 50 000. Accept a clear comparison. |
| 6 | The claim is wrong. 12 months: 24 671 coins; 24 months: 50 864 coins, which is more than 2 × 24 671 = 49 342 by about 1522 coins. Interest earns interest, so more than twice as much. Without interest it would be exactly double: 24 000 and 48 000. |
| 7 | Assumptions: the rate stays 0.5% every month; every payment is made on time and in full; nothing is withdrawn; interest is added at month end; the loom still costs 50 000 coins. The rate and a missed payment are the most likely to fail. Credit any reasoned choice, and a stated effect on the answer. |
| 8 (criterion C) | The note gives the number of months, the total at that point, the effect of a lower rate and one assumption, with units, in sentences that make sense without the working. |
Exemplar answers
| Task D1 response | What it looks like |
|---|---|
| Level 3–4 (D) | “After 3 months 6030 coins. The formula is 400 000(1.005^n − 1). Putting n = 24 gives 50 864 so it takes 24 months. Without interest it is 25 months. At 0.3% it is 25 months.” Correct answers, but no comparison of 23 months, the claim is not tested and nothing is said about assumptions. |
| Level 7–8 (D and C) | Both 23 and 24 months shown (48 621 and 50 864) to justify 24; 0.3% gives 49 693 then 51 842, so 25 months, the same as no interest, and the group should look for a better rate; “12 months gives 24 671, so the double of that, 49 342, is less than 24 months at 50 864, because interest earns interest. The model assumes the rate never changes, which a bank is not bound to do. If it falls, the month in which the target is reached goes from 24 to 25.” The note to the treasurer is four clear sentences with units. |
Differentiation
A way in for anyone stuck
- Give a table with columns for the month number, the payment, the number of months of interest and the amount for Part 1.
- Allow trial and improvement with a calculator for question 3 instead of solving 1.005n > 1.125.
- For Part 4, ask: “If the bank changed the rate half way, what would you do about it?”
Stretch
- The group wants 50 000 coins in 12 months. What monthly payment, to the nearest coins, would be needed at 0.5%? (4053 coins.)
- Without interest, what monthly payment reaches 50 000 coins in 12 months? How much does the interest save each month? (4167 coins; interest saves about 113 coins a month.)
Students learning in English as an additional language
| Word in task D1 | Plain meaning |
|---|---|
| interest | extra money the bank adds to your savings |
| payment | money paid into the account |
| total | the whole amount in the account |
| target | the amount you want to reach |
| assume | decide to treat something as true to make a model |
Sentence starters for task D1:
- After … months the total is … because …
- The target is first reached after … months, since …
- I recommend … because …
Task D2: The clinic dosage plan
Why set it. Geometric series, rearranging a formula and direct variation inside one real decision. An alternative to D1: set one of them.
When: End of unit, after lesson 7. Time: 50 minutes. Format: Individual, in class, one lesson. A written solution with explanations and a short reply to the head nurse.
The student sheet, with its task-specific clarification
When: End of unit, after lesson 7. Time: 50 minutes. Calculator: scientific calculator. Criteria: D, applying mathematics in real-life contexts; C, communicating.
A clinic in Phuket gives a medicine to a patient every 8 hours. A standard dose for an adult of 60 kg is 150 mg. By the time of the next dose, 60% of the previous amount is still in the patient’s body. The clinic’s safe limit for the level of the medicine in the body is 350 mg.
Part 1. Doses add up
- Find the level in the body just after the 1st, 2nd and 3rd doses.
- Find the long-run level the patient approaches, just after a dose. Explain why the level never quite reaches it.
- Which dose is the first to take the level above the safe limit of 350 mg? Show the two levels that decide it.
Part 2. Adjusting the dose
- The long-run level L is related to the dose D and the fraction f that remains by L = D ÷ (1 − f). Make D the subject. What is the largest dose that keeps the long-run level from going above 350 mg?
- A dose varies directly with body mass. What dose should a 40 kg child receive if the same fraction remains? What long-run level results?
Part 3. A patient who clears the medicine slowly
- A 40 kg child clears the medicine more slowly: 80% remains at the next dose. Is the child’s dose from question 5 still safe in the long run? What dose would keep the long-run level at 350 mg?
- Show that halving the dose and halving the amount that leaves (1 − f) leaves the long-run level exactly the same. Why does this matter for a patient with slow clearance?
Part 4. Does it make sense?
- State what the model assumes, and say what could make the real level different from your answers.
- Write a short reply to the head nurse with your advice, so that someone who has not seen your working can understand it.
How your work is judged
| Level | Criterion D in task D2: what it looks like |
|---|---|
| 1–2 | You find some of the levels correctly, for example after the first doses, without the long-run level or a decision. |
| 3–4 | You find the levels, the long-run level and the first dose over the limit, and say whether the plan is safe. |
| 5–6 | You use suitable methods correctly for the levels, the rearranged formula, the proportional dose and the slow-clearance case, and say how reliable the model is. |
| 7–8 | You choose and use suitable methods throughout, reach correct answers, explain what the model leaves out and how that affects safety, and justify your advice. |
| Level | Criterion C in task D2: what it looks like |
|---|---|
| 1–2 | Some working is shown, but it is hard to see how the levels or the doses were reached. |
| 3–4 | Your working can be followed, with some correct notation and units in mg. |
| 5–6 | Your working is clear, usually in order with units and notation, and your reply to the head nurse is mostly well organized. |
| 7–8 | Your working is clear, complete and well ordered, units and notation are always correct, and your reply makes sense on its own. |
Markscheme
| Task D2 question | Answer and marking |
|---|---|
| 1 | 150, 240, 294 mg. 150; 150 + 150 × 0.6 = 240; 150(1 + 0.6 + 0.36) = 294. 1 + 0.6 + 0.36 = 1.96. A list of 150, 300, 450 ignores what leaves; 210 and 234 for the 2nd and 3rd levels come from keeping 40% in place of 60%. |
| 2 | 375 mg. 150 ÷ (1 − 0.6) = 150 ÷ 0.4. The level after n doses is 375(1 − 0.6n); 0.6n is never 0, so it stays just under 375. |
| 3 | The 6th dose. After 5 doses 375(1 − 0.65) = 345.84 mg, below 350; after 6 doses 375(1 − 0.66) = 357.50 mg, above. Solving 0.6n < 1/15 gives n > 5.3, so 6. |
| 4 | D = L(1 − f); 140 mg. 350 × 0.4 = 140 mg. The long-run level is exactly 350 at 140 mg, which is never passed. |
| 5 | 100 mg; long-run level 250 mg. 150 ÷ 60 = 2.5 mg per kg, 2.5 × 40 = 100 mg; 100 ÷ 0.4 = 250 mg. The level is also 40/60 of 375, which is 250. |
| 6 | No. 100 ÷ (1 − 0.8) = 500 mg, above 350 mg. For a long-run level of 350 mg: D = 350 × 0.2 = 70 mg. The slower clearance means the same dose builds up to twice the level: 500 mg against 250 mg. |
| 7 | The level is unchanged. (D/2) ÷ ((1 − f)/2) = D ÷ (1 − f). For example 75 ÷ 0.2 = 375 and 150 ÷ 0.4 = 375. The long-run level depends on the ratio, so a patient who clears the medicine half as fast needs half the dose to stay at the same level. |
| 8 | Assumptions: a fixed fraction remains every time; doses are exactly 8 hours apart; no other medicine interferes; the safe limit is the same for everybody. Real patients differ, so the level should be checked by a clinician. Credit a reasoned, specific limitation. |
| 9 (criterion C) | The reply states the dose, the long-run level, the first dose over the limit (for the standard plan), the adjustment for slow clearance, with units, in sentences that make sense without the working. |
Exemplar answers
| Task D2 response | What it looks like |
|---|---|
| Level 3–4 (D) | “After the three doses the levels are 150, 240 and 294 mg. The long-run level is 150 ÷ 0.4 = 375 mg. The limit is passed at the 6th dose. A child gets 100 mg. The slow child has 500 mg which is too high.” Correct answers, but the limiting dose is not shown with two levels, the safe dose for the slow child is not found and the assumptions are not discussed. |
| Level 7–8 (D and C) | “The levels 345.84 mg and 357.50 mg show that the 6th dose is the first over 350 mg. D = L(1 − f), so the largest dose is 140 mg. The child gets 100 mg by mass. If 80% remains, the ceiling is 100 ÷ 0.2 = 500 mg, so the dose must be 70 mg. Halving the dose and halving the amount that leaves gives the same ceiling, so a slow-clearing patient needs a proportionally smaller dose. The model assumes the same fraction every time, which a real patient may not follow.” The reply to the head nurse is four clear sentences with units. |
Differentiation
A way in for anyone stuck
- Give a table with columns for the dose number, the amount left from before and the level after the dose for Part 1.
- Allow a stepwise calculation with a calculator for question 3 instead of solving 0.6n < 1/15.
- For question 7, ask: “What does the long-run level depend on, the dose, the fraction that leaves, or both?”
Stretch
- The adult dose is given every 12 hours and 36% remains. What is the long-run level? Is it safer than the 8-hour plan? (150 ÷ 0.64 = 234.4 mg; it is lower.)
- A first dose of 250 mg is followed by 140 mg each time. Find the levels after the 2nd and 3rd doses and the long-run level. (290, 314 and 350 mg: it approaches 350 mg from below.)
Students learning in English as an additional language
| Word in task D2 | Plain meaning |
|---|---|
| dose | one measured amount of medicine |
| remains | is still there |
| long-run level | the number the level gets close to after many doses |
| safe limit | the highest level that is safe |
| clear | leave the body (a drug clears from the body) |
Sentence starters for task D2:
- Just after the … dose the level is … because …
- The long-run level is …, which is / is not above the limit, so …
- I advise … because …
Practicalities
Every task is original and written for this unit; none is copied or adapted from a published book.
The student sheets, with their task-specific clarifications, are public on the unit page so students know what each level looks like before they start. Markschemes and exemplars are only on this page.
To print a sheet, open it on the unit page, then print the page. A1 and C1 need no calculator; B1, A2, D1 and D2 use a scientific calculator, and a GDC is optional for any of them.
The sheets are public, so a student may have seen A1 and A2. For an unseen A2, use this parallel version (same marks, same markscheme notes): Q1 5, 10, 20, … to 8 terms (1275). Q2 24, 8, 8/3, … to infinity (36). Q3 0.6363… as a fraction (7/11). Q6 first term 4, ratio 1.5, first n with a sum above 200 (n = 9: S8 = 197.03 and S9 = 299.55). Q9 sum to infinity 18 and sum of the first two terms 16 (1 − r2 = 8/9: a = 12, r = 1/3, or a = 24, r = −1/3). Q4 make a the subject of S = n/2 (a + l) (a = 2S/n − l; test n = 10, l = 39, S = 210 gives a = 3). Q5 y varies inversely as x2, y = 8 when x = 3, find y when x = 6 (k = 72, y = 2). Q7 third term 14 and eighth term 39, sum of the first 20 terms (d = 5, a = 4, 1030). Q8 show Σk=1n (6k − 1) = 3n2 + 2n, then find n for a sum of 320 ((3n + 32)(n − 10) = 0, n = 10). Q10 Eli: “Doubling the common difference of an arithmetic series doubles its sum.” (1 + 2 + 3 + 4 + 5 = 15 but 1 + 3 + 5 + 7 + 9 = 25; true only when the first term is 0.) Q11 Q varies as the square of r and the cube of h; r rises 10% and h falls 20% (1.21 × 0.512 = 0.61952, down about 38%). For A1, change the numbers the same way.
The Year 5 descriptor bands on this page are band summaries written for this site; check them against the current IB MYP Mathematics guide before relying on them for reporting.