The shape of the lesson
By the end of the lesson, learners can
- The common difference: In the lab, set the first term and the common difference with the sliders and watch each term grow by the same step.
- Summing an AP: Write the sum forwards and backwards and add the two lines, as in Fig. 1.
- Problems in context: Wages that rise by a fixed amount, savings that grow by a fixed sum, rows of seats that increase by two: each is an AP.
- Identify: \(a\) is the first term, \(d\) the constant step (it can be negative).
- One term: \(u_n = a + (n-1)d\).
- Sum: \(S_n = \dfrac{n}{2}\left[2a + (n-1)d\right]\) or \(\dfrac{n}{2}(a + l)\).
- Unknowns: two facts give two simultaneous equations in \(a\) and \(d\); three terms in AP give \(2b = a + c\).
- Inequalities: solve the quadratic, then take the next whole number and check both sides.
One picture every idea returns to
An AP with first term \(a\) and common difference \(d\) has \(n\)th term \(u_n = a + (n-1)d\) and sum of the first \(n\) terms \(S_n = \dfrac{n}{2}\left[2a + (n-1)d\right] = \dfrac{n}{2}(a + l)\), where \(l\) is the last term. Three numbers \(a, b, c\) are in AP when \(2b = a + c\).
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Add the same amount, every time.. An arithmetic progression grows by a fixed step, the common difference. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: The common difference | In the lab, set the first term and the common difference with the sliders and watch each term grow by the same step. Project tab 2 of the lab (Sequences and proportion lab, linear sequences) and let a student drive it while the class predicts. Model the worked example A term, and which term on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Summing an AP | Write the sum forwards and backwards and add the two lines, as in Fig. 1. Model the worked example A straight sum on the board, one line at a time. Your turn: The sum of the first \(n\) terms of a progression is \(S_n = 3n^2 + 2n\). Find the first term and the common difference. | Pairs |
| 21 to 30 min | Teach and check: Problems in context | Wages that rise by a fixed amount, savings that grow by a fixed sum, rows of seats that increase by two: each is an AP. Model the worked example A savings plan on the board, one line at a time. | Rally Coach |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 25 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Problems in context, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Rally Coach
Pairs share one worksheet. One partner solves the first question aloud while the other coaches, checking each line and offering a prompt rather than the answer; they swap roles for the next question.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| The 20th term is \(a + 19d\), not \(a + 20d\): the first term already counts as one of the twenty. | Return to The common difference: ask a pair to state the correct version and show it on the lab before moving on. |
| Counting terms between two numbers: \(\dfrac{297 - 102}{3} = 65\) counts the gaps; there is one more term than gaps, so \(n = 66\). | Return to Summing an AP: ask a pair to state the correct version and show it on the lab before moving on. |
| Read whether the question asks for one term or the total. 'How much in the 24th month' is \(u_{24}\); 'how much in the first 24 months' is \(S_{24}\). | Return to Problems in context: ask a pair to state the correct version and show it on the lab before moving on. |
Arithmetic progressions: worksheet
6 questions, 25 marks. Written in the style of the papers; no past paper question is reproduced.
For the AP \(15, 11, 7, \ldots\) find (a) the 25th term, (b) the sum of the first 25 terms.
The 4th term of an AP is 19 and the sum of the first 10 terms is 235. Find the first term and the common difference.
Prove that the sum of the first \(n\) terms of an arithmetic progression with first term \(a\) and common difference \(d\) is \(\dfrac{n}{2}\left[2a + (n-1)d\right]\).
The numbers \(k\), \(3k - 2\) and \(4k + 3\) are the first three terms of an arithmetic progression.
(a) Find \(k\) and the common difference.
(b) Find the sum of the first 15 terms.
The sum of the first \(n\) terms of a progression is \(S_n = 4n^2 - n\).
(a) Find the first two terms.
(b) Show that the progression is arithmetic and find an expression for the \(n\)th term.
A runner trains every day. She runs 2 km on day 1 and increases her distance by 0.25 km each day.
(a) Find the distance she runs on day 20.
(b) Find the total distance she runs in the first 20 days.
(c) Find the day on which her total distance first exceeds 200 km.
Answer key and marking notes
- \(a = 15\), \(d = -4\).
- (a) \(u_{25} = 15 + 24 \times (-4) = -81\)
- (b) \(S_{25} = \dfrac{25}{2}(15 + (-81)) = 12.5 \times (-66) = -825\)
- \(a + 3d = 19\), so \(2a + 6d = 38\).
- \(\dfrac{10}{2}(2a + 9d) = 235\), so \(2a + 9d = 47\).
- Subtract: \(3d = 9\), \(d = 3\); then \(a = 19 - 9 = 10\).
- \(S_n = a + (a + d) + \cdots + \left[a + (n-1)d\right]\)
- Reversed: \(S_n = \left[a + (n-1)d\right] + \left[a + (n-2)d\right] + \cdots + a\)
- Adding term by term, each of the \(n\) pairs sums to \(2a + (n-1)d\), so \(2S_n = n\left[2a + (n-1)d\right]\).
- Hence \(S_n = \dfrac{n}{2}\left[2a + (n-1)d\right]\).
- (a) \(2(3k - 2) = k + 4k + 3\), so \(6k - 4 = 5k + 3\) and \(k = 7\).
- The terms are \(7, 19, 31\), so \(d = 12\).
- (b) \(S_{15} = \dfrac{15}{2}\left[14 + 14 \times 12\right] = 7.5 \times 182 = 1365\)
- (a) \(u_1 = S_1 = 3\); \(S_2 = 14\), so \(u_2 = 11\).
- (b) \(u_n = S_n - S_{n-1} = (4n^2 - n) - \left[4(n-1)^2 - (n-1)\right] = 4n^2 - n - 4n^2 + 8n - 4 + n - 1 = 8n - 5\).
- \(u_n - u_{n-1} = 8\), a constant, so the progression is arithmetic with \(d = 8\) and \(u_n = 8n - 5\).
- (a) \(u_{20} = 2 + 19 \times 0.25 = 6.75\) km
- (b) \(S_{20} = \dfrac{20}{2}(2 + 6.75) = 87.5\) km
- (c) \(S_n = \dfrac{n}{2}\left[4 + 0.25(n-1)\right] = \dfrac{n(0.25n + 3.75)}{2} > 200\), so \(n^2 + 15n - 1600 > 0\).
- Positive root \(n = \dfrac{-15 + \sqrt{6625}}{2} = 33.2\), so \(n = 34\).
- Check: \(S_{33} = 198\) km and \(S_{34} = 208.25\) km, so day 34.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
Find the 15th term of the arithmetic progression \(40, 37, 34, \ldots\)
- A\(-2\)
- B\(-5\)
- C\(82\)
- D\(1\)
Find the sum of the first 25 terms of the arithmetic progression \(3, 7, 11, \ldots\)
- A1275
- B1325
- C1237.5
- D2550
An arithmetic progression has first term 6 and 10th term 33. Find the common difference.
- A3
- B2.7
- C3.3
- D27
How many terms are there in the arithmetic progression \(12, 17, 22, \ldots, 287\)?
- A56
- B55
- C57
- D57.4
The sum of the first \(n\) terms of an arithmetic progression is \(S_n = 3n^2 - n\). Find the common difference.
- A6
- B3
- C8
- D10
Find the sum of all the integers from 1 to 200 inclusive that are not multiples of 4.
- A15 000
- B5100
- C15 200
- D20 100
The third term of an arithmetic progression is 14 and the sum of the first 10 terms is 215. Find the first term \(a\) and the common difference \(d\).
- A\(a = 8,\ d = 3\)
- B\(a = 9,\ d = 2.5\)
- C\(a = -1,\ d = 5\)
- D\(a = 2,\ d = 6\)
Logs are stacked in rows. The bottom row has 20 logs and each row has one log fewer than the row below it. The stack contains 155 logs in total. How many rows are there?
- A10
- B31
- C10 or 31
- D8
Quiz key
\(a = 40\), \(d = -3\), so \(u_{15} = 40 + 14 \times (-3) = -2\). The tempting \(-5\) uses \(15d\) instead of \(14d\); the first term already has no \(d\) added. 82 uses \(d = +3\), and 1 uses \(13d\).
\(S_{25} = \dfrac{25}{2}\left[2 \times 3 + 24 \times 4\right] = \dfrac{25}{2} \times 102 = 1275\). The tempting 1325 uses \(25d\) instead of \(24d\). 1237.5 uses \(a\) instead of \(2a\) in the bracket (a sum of whole numbers cannot be a decimal), and 2550 forgets to halve.
\(6 + 9d = 33\), so \(9d = 27\) and \(d = 3\). The tempting 2.7 divides 27 by 10, but there are only 9 steps from the 1st term to the 10th. 3.3 is \(33 \div 10\), ignoring the first term, and 27 is the total increase, not one step.
\((n - 1) \times 5 = 275\), so \(n - 1 = 55\) and \(n = 56\). The tempting 55 counts the gaps between terms rather than the terms (a fencepost error). 57.4 is \(287 \div 5\), which ignores the first term 12, and 57 rounds that wrong value.
\(S_1 = 2\), so \(u_1 = 2\). \(S_2 = 12 - 2 = 10\), so \(u_2 = 10 - 2 = 8\). Then \(d = 8 - 2 = 6\). The tempting 3 reads off the coefficient of \(n^2\); comparing with \(S_n = \dfrac{d}{2}n^2 + \cdots\), that coefficient is \(\dfrac{d}{2}\), not \(d\). 8 is the second term and 10 is \(S_2\).
\(1 + 2 + \cdots + 200 = \dfrac{200}{2}(1 + 200) = 20\,100\). The multiples of 4 are \(4, 8, \ldots, 200\): 50 terms, sum \(\dfrac{50}{2}(4 + 200) = 5100\). Answer \(20\,100 - 5100 = 15\,000\). The tempting 15 200 stops the multiples at 196 (49 terms, sum 4900) and so misses 200. 5100 is the sum of the multiples themselves, and 20 100 is the sum of every integer.
\(a + 2d = 14\) gives \(2a + 4d = 28\), and \(5(2a + 9d) = 215\) gives \(2a + 9d = 43\). Subtracting, \(5d = 15\), so \(d = 3\) and \(a = 8\). The tempting \(a = 9, d = 2.5\) uses \(10d\) instead of \(9d\) in the sum. \(a = -1, d = 5\) takes the third term as \(a + 3d\), and \(a = 2, d = 6\) fits the third term but gives a sum of 290.
\(\dfrac{n}{2}\left[40 - (n - 1)\right] = 155\) gives \(n(41 - n) = 310\), so \(n^2 - 41n + 310 = 0\), \((n - 10)(n - 31) = 0\). With 31 rows the top row would have \(20 - 30 = -10\) logs, which is impossible, so there are 10 rows (top row 11 logs). The tempting "10 or 31" and 31 accept the algebra without checking the context. 8 rows hold only \(\dfrac{8}{2}(20 + 13) = 132\) logs.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.