The shape of the lesson
By the end of the lesson, learners can
- The common ratio: In the lab, choose a first term and a multiplier and watch the terms.
- The sum of a GP: Write \(S_n = a + ar + \cdots + ar^{n-1}\), multiply every term by \(r\), and subtract.
- Sum to infinity: If \(|r| < 1\), then \(r^n \to 0\) as \(n\) grows, so \(S_n = \dfrac{a(1 - r^n)}{1 - r}\) gets as close as we like to \(\dfrac{a}{1 - r}\).
- More than one progression: Paper 1 questions often link an AP and a GP: some terms of one are terms of the other.
- Ratio: \(r = \dfrac{u_{n+1}}{u_n}\); two given terms: divide to eliminate \(a\).
- One term: \(u_n = ar^{n-1}\).
- Sum: \(S_n = \dfrac{a(1 - r^n)}{1 - r}\), or \(\dfrac{a(r^n - 1)}{r - 1}\) when \(r > 1\).
- Sum to infinity: only if \(|r| < 1\); then \(S_\infty = \dfrac{a}{1 - r}\).
- Two progressions: write the shared terms with \(a, d\) or \(a, r\) and use \(2b = a + c\) or \(b^2 = ac\).
One picture every idea returns to
A GP with first term \(a\) and common ratio \(r\) has \(u_n = ar^{n-1}\) and \(S_n = \dfrac{a(1 - r^n)}{1 - r}\) (\(r \ne 1\)). It converges when \(|r| < 1\), and then \(S_\infty = \dfrac{a}{1 - r}\). Three numbers \(a, b, c\) are in GP when \(b^2 = ac\).
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, Multiply, and sometimes reach infinity.. A geometric progression multiplies by the same ratio each time. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 11 min | Teach and check: The common ratio | In the lab, choose a first term and a multiplier and watch the terms. Project tab 4 of the lab (Sequences and proportion lab, exponential sequences) and let a student drive it while the class predicts. Model the worked example A term on the board, one line at a time. | Whole class, then pairs |
| 11 to 17 min | Teach and check: The sum of a GP | Write \(S_n = a + ar + \cdots + ar^{n-1}\), multiply every term by \(r\), and subtract. Model the worked example A growing GP on the board, one line at a time. | Pairs |
| 17 to 23 min | Teach and check: Sum to infinity | If \(|r| < 1\), then \(r^n \to 0\) as \(n\) grows, so \(S_n = \dfrac{a(1 - r^n)}{1 - r}\) gets as close as we like to \(\dfrac{a}{1 - r}\). Model the worked example Sum to infinity and a recurring decimal on the board, one line at a time. Your turn: Find the set of values of \(x\) for which the GP \(1 + 2x + 4x^2 + \cdots\) has a sum to infinity, and find that sum when \(x = 0.2\). | Pairs |
| 23 to 30 min | Teach and check: More than one progression | Paper 1 questions often link an AP and a GP: some terms of one are terms of the other. Model the worked example AP terms that form a GP on the board, one line at a time. | Numbered Heads Together |
| 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, 26 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 More than one progression, 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.
Numbered Heads Together
Teams of four number off. Put up a question, give the team time to make sure every member can answer, then call a number: that student answers for the team, so nobody can leave the thinking to someone else.
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 |
|---|---|
| Dividing \(ar^5\) by \(ar^2\) gives \(r^3\), not \(r^4\): subtract the powers, \(5 - 2 = 3\). | Return to The common ratio: ask a pair to state the correct version and show it on the lab before moving on. |
| With a negative ratio, bracket it: \(\left(-\tfrac{1}{3}\right)^6 = +\dfrac{1}{729}\), and \(1 - r = 1 - \left(-\tfrac{1}{3}\right) = \tfrac{4}{3}\), not \(\tfrac{2}{3}\). | Return to The sum of a GP: ask a pair to state the correct version and show it on the lab before moving on. |
| \(S_\infty\) only exists when \(|r| < 1\). | A GP such as \(2 - 4 + 8 - \cdots\) has \(r = -2\), so \(\dfrac{a}{1 - r} = \dfrac{2}{3}\) is meaningless. |
| When an equation like \(d^2 = 5d\) appears, do not divide by \(d\) blindly: say \(d \ne 0\) (given), so \(d = 5\). | If \(d = 0\) were allowed it would also be a solution. |
Geometric progressions and sum to infinity: worksheet
6 questions, 26 marks. Written in the style of the papers; no past paper question is reproduced.
For the GP \(5, 15, 45, \ldots\) find (a) the 8th term, (b) the sum of the first 8 terms.
The 3rd term of a GP is 12 and the 6th term is \(-96\). Find the first term, the common ratio and the sum of the first 10 terms.
(a) Prove that the sum of the first \(n\) terms of a GP with first term \(a\) and common ratio \(r\) (\(r \ne 1\)) is \(\dfrac{a(1 - r^n)}{1 - r}\).
(b) Deduce the sum to infinity when \(|r| < 1\), explaining why the condition is needed.
(a) Use a geometric progression to express the recurring decimal \(0.363636\ldots\) as a fraction in its lowest terms.
(b) Find the sum to infinity of the GP \(18 - 12 + 8 - \cdots\)
An arithmetic progression has first term 9 and common difference \(d\), where \(d \ne 0\). The 1st, 4th and 13th terms of the AP are the first three terms of a geometric progression.
(a) Find \(d\).
(b) Find the common ratio of the GP.
(c) Find the sum of the first 20 terms of the AP.
A ball is dropped from a height of 10 m. After each bounce it rises to 60% of the height from which it last fell.
(a) Find the height the ball reaches after the 4th bounce.
(b) Find the total distance travelled by the ball up to the moment it hits the ground for the 5th time.
(c) Find the total distance the ball travels before it comes to rest.
Answer key and marking notes
- \(a = 5\), \(r = 3\).
- (a) \(u_8 = 5 \times 3^7 = 10\,935\)
- (b) \(S_8 = \dfrac{5(3^8 - 1)}{3 - 1} = \dfrac{5 \times 6560}{2} = 16\,400\)
- \(ar^2 = 12\), \(ar^5 = -96\), so \(r^3 = -8\) and \(r = -2\).
- \(a = \dfrac{12}{4} = 3\).
- \(S_{10} = \dfrac{3\left(1 - (-2)^{10}\right)}{1 - (-2)} = \dfrac{3(1 - 1024)}{3} = -1023\)
- (a) \(S_n = a + ar + ar^2 + \cdots + ar^{n-1}\) and \(rS_n = ar + ar^2 + \cdots + ar^{n}\).
- Subtract: \(S_n - rS_n = a - ar^n\), so \((1 - r)S_n = a(1 - r^n)\) and \(S_n = \dfrac{a(1 - r^n)}{1 - r}\).
- (b) If \(|r| < 1\), \(r^n \to 0\) as \(n \to \infty\), so \(S_n \to \dfrac{a}{1 - r}\).
- If \(|r| \ge 1\), \(r^n\) does not tend to 0, so \(S_n\) has no finite limit.
- (a) \(a = 0.36\), \(r = 0.01\): \(S_\infty = \dfrac{0.36}{0.99} = \dfrac{36}{99} = \dfrac{4}{11}\)
- (b) \(r = -\dfrac{12}{18} = -\dfrac{2}{3}\), \(|r| < 1\).
- \(S_\infty = \dfrac{18}{1 + \frac{2}{3}} = \dfrac{18}{5/3} = \dfrac{54}{5} = 10.8\)
- (a) The terms are \(9\), \(9 + 3d\), \(9 + 12d\). In GP: \((9 + 3d)^2 = 9(9 + 12d)\).
- \(81 + 54d + 9d^2 = 81 + 108d\), so \(9d^2 = 54d\); \(d \ne 0\), so \(d = 6\).
- (b) The terms are \(9, 27, 81\), so \(r = 3\).
- (c) \(S_{20} = \dfrac{20}{2}\left[18 + 19 \times 6\right] = 10 \times 132 = 1320\)
- (a) Heights after each bounce form a GP: \(6, 3.6, 2.16, \ldots\), so after the 4th bounce \(10 \times 0.6^4 = 1.296 = 1.30\) m (3 s.f.).
- (b) The first fall is 10 m; after each of the first 4 bounces the ball goes up and down the same height.
- \(10 + 2(6 + 3.6 + 2.16 + 1.296) = 10 + 2 \times 13.056 = 36.1\) m (3 s.f.)
- (c) The rises form a GP with \(a = 6\), \(r = 0.6\): \(S_\infty = \dfrac{6}{0.4} = 15\).
- Total \(= 10 + 2 \times 15 = 40\) m
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 7th term of the geometric progression \(5, -10, 20, \ldots\)
- A320
- B\(-320\)
- C\(-640\)
- D640
Find the sum to infinity of the geometric progression \(18 - 6 + 2 - \cdots\)
- A13.5
- B27
- C9
- DThere is no sum to infinity
Find the sum of the first 6 terms of the geometric progression \(2, 6, 18, \ldots\)
- A728
- B2186
- C364
- D1456
For which of these values of \(x\) does the geometric progression \(1 + 3x + 9x^2 + \cdots\) have a sum to infinity?
- A\(x = 0.3\)
- B\(x = 0.4\)
- C\(x = -0.5\)
- D\(x = \dfrac{1}{3}\)
The sum of the first two terms of a geometric progression is 15 and its sum to infinity is 27. The common ratio \(r\) is positive. Find \(r\).
- A\(\dfrac{2}{3}\)
- B\(-\dfrac{2}{3}\)
- C\(\dfrac{4}{9}\)
- D\(\dfrac{5}{9}\)
Find the least number of terms of the geometric progression \(3 + 6 + 12 + \cdots\) that must be taken for the sum to exceed 5000.
- A11
- B10
- C12
- D9
An arithmetic progression has first term \(a\) and common difference \(d\), where \(d \ne 0\). Its 1st, 3rd and 9th terms are the first three terms of a geometric progression. Find the common ratio of the geometric progression.
- A3
- B9
- C\(\dfrac{1}{3}\)
- D5
Mina's salary is $30 000 in her first year of work. It increases by 4% every year. Find the total she earns in her first 10 years, to the nearest dollar.
- A$360 183
- B$354 000
- C$404 591
- D$444 073
Quiz key
\(a = 5\), \(r = -2\), so \(u_7 = 5 \times (-2)^6 = 5 \times 64 = 320\). The tempting \(-320\) assumes the minus sign survives, but an even power of \(-2\) is positive (odd-numbered terms are positive here). \(\pm 640\) use \(r^7\) instead of \(r^6\).
\(r = \dfrac{-6}{18} = -\tfrac{1}{3}\), and \(|r| < 1\), so \(S_\infty = \dfrac{18}{1 - (-\tfrac{1}{3})} = \dfrac{18}{4/3} = 13.5\). The tempting 27 uses \(r = +\tfrac{1}{3}\), ignoring the alternating signs. 9 is \(\dfrac{18}{2}\), and a negative ratio does not stop a sum to infinity existing; only \(|r| \ge 1\) does.
\(S_6 = \dfrac{2(3^6 - 1)}{3 - 1} = \dfrac{2 \times 728}{2} = 728\). Check: \(2 + 6 + 18 + 54 + 162 + 486 = 728\). The tempting 2186 uses \(3^7\), one term too many. 364 leaves out \(a = 2\), and 1456 forgets to divide by \(r - 1 = 2\).
\(r = 3x\), so we need \(|3x| < 1\), that is \(-\tfrac{1}{3} < x < \tfrac{1}{3}\). Only \(x = 0.3\) works (\(r = 0.9\)). The tempting \(x = \tfrac{1}{3}\) gives \(r = 1\), and the strict inequality excludes it: the terms would all be 1 and the sum would grow without limit. \(x = 0.4\) gives \(r = 1.2\) and \(x = -0.5\) gives \(r = -1.5\).
\(27(1 - r)(1 + r) = 15\), so \(1 - r^2 = \dfrac{15}{27} = \dfrac{5}{9}\) and \(r^2 = \dfrac{4}{9}\). As \(r > 0\), \(r = \dfrac{2}{3}\) (then \(a = 9\): \(9 + 6 = 15\) and \(\dfrac{9}{1/3} = 27\)). The tempting \(\dfrac{4}{9}\) stops at \(r^2\). \(-\dfrac{2}{3}\) is the other root, ruled out by \(r > 0\), and \(\dfrac{5}{9}\) is \(1 - r^2\).
\(3(2^n - 1) > 5000\) gives \(2^n > 1667.7\). Since \(2^{10} = 1024\) and \(2^{11} = 2048\), \(n = 11\). Check: \(S_{10} = 3069\) and \(S_{11} = 6141\). The tempting 10 stops one term early: \(S_{10} = 3069\) is still below 5000. 12 is the first term above 5000 (\(u_{12} = 3 \times 2^{11} = 6144\)), not the first sum, and 9 gives a sum of only 1533.
\((a + 2d)^2 = a(a + 8d)\) gives \(a^2 + 4ad + 4d^2 = a^2 + 8ad\), so \(4d^2 = 4ad\). As \(d \ne 0\), \(d = a\). The terms are \(a, 3a, 9a\), so the ratio is 3. The tempting 9 is the ratio of the 3rd GP term to the 1st, which is \(r^2\). \(\dfrac{1}{3}\) divides each term by the next one, and 5 comes from expanding \((a + 2d)^2\) as \(a^2 + 4d^2\), which gives \(d = 2a\) and terms \(a, 5a, 17a\) that are not even geometric.
\(S_{10} = \dfrac{30\,000(1.04^{10} - 1)}{1.04 - 1} = 360\,183.21\ldots\), so $360 183. The tempting $354 000 adds a fixed 4% of $30 000 ($1200) each year, an AP; each rise is 4% of a larger salary, so the true total is higher. $404 591 sums 11 years, and $444 073 is 10 times the salary \(30\,000 \times 1.04^{10}\), which is the 11th year's pay.
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.