中文（繁體）
中文（简体）
English
S5 Maths | Arithmetic and Geometric Sequences prep guide (for tutors) | terms, sums and sum to infinity
Students can quote both formulas, then swap a and d the moment a question asks for the nth term, or use the infinite sum without checking |r| < 1. One timeline separates the two ideas.
Objectives — Students can: (1) write the general term from the first term and common difference or ratio, and recover a and d from two terms; (2) choose the right sum formula and handle the number-of-terms trap; (3) state when an infinite geometric sum exists and justify it with |r| < 1.
01 Prerequisites (3-minute check)
If any of these fails, fix it before the lesson starts.
| Skill | Quick check | Pass mark |
|---|---|---|
| Solving a linear equation | 3 + 9d = 39, find d | 4 |
| Index laws | value of 3^5 | 243 |
| Inequalities | meaning of |r| < 1 | -1 < r < 1 |
02 Lesson flow (60 minutes)
| Time | Stage | Tutor does | Student does |
|---|---|---|---|
| 0-10 min | Sequence language | a, d, r, n, T(n), S(n) | write the first four terms |
| 10-24 min | Arithmetic term and sum | a = 3, d = 4 | T(10) = 39, S = 210 |
| 24-38 min | Geometric term and sum | a = 2, r = 3 | T(6) = 486, S = 728 |
| 38-48 min | Sum to infinity | a = 8, r = 0.5 | S = 16 |
| 48-56 min | Applications and counting terms | multiples of 3 up to 99 | n = 33, S = 1683 |
| 56-60 min | Exit check | three questions, marked at once | independent |
03 Teaching script (read it out)
Scene 1: Arithmetic: a straight line, with a trapezium for the sum
a = 3, d = 4 gives 3, 7, 11, 15 …
T(n) = a + (n – 1)d, so T(10) = 3 + 9 x 4 = 39. The (n – 1) matters: the first term adds no d. Substituting n = 1 is the fastest self-check, since T(1) must equal a.
S(n) = n/2 [2a + (n – 1)d] = 10/2 x [6 + 36] = 210. The n/2 (first + last) form is the same formula; students should be able to say why.
Reverse problem: T(4) = 15 and T(9) = 35 give 5d = 20, so d = 4 and a = 3. Subtracting the two equations is the only reliable method.
T(n) = a + (n – 1)d, so T(10) = 3 + 9 x 4 = 39. The (n – 1) matters: the first term adds no d. Substituting n = 1 is the fastest self-check, since T(1) must equal a.
S(n) = n/2 [2a + (n – 1)d] = 10/2 x [6 + 36] = 210. The n/2 (first + last) form is the same formula; students should be able to say why.
Reverse problem: T(4) = 15 and T(9) = 35 give 5d = 20, so d = 4 and a = 3. Subtracting the two equations is the only reliable method.
Task: a sequence has T(3) = 11 and T(8) = 31. Find a and d (3 and 4) and then T(20) (79).
Scene 2: Geometric: terms multiply, finite sums divide
a = 2, r = 3 gives 2, 6, 18, 54 …
T(n) = ar^(n-1), so T(6) = 2 x 3^5 = 486.
S(n) = a(r^n – 1)/(r – 1) = 2 x (3^6 – 1)/2 = 728. The r = 1 case needs separate treatment because the denominator vanishes; then S = na. Almost everyone forgets it.
Geometric growth dwarfs arithmetic: with a = 2, term 6 is 12 by adding and 486 by multiplying. That comparison explains compound interest better than a definition.
T(n) = ar^(n-1), so T(6) = 2 x 3^5 = 486.
S(n) = a(r^n – 1)/(r – 1) = 2 x (3^6 – 1)/2 = 728. The r = 1 case needs separate treatment because the denominator vanishes; then S = na. Almost everyone forgets it.
Geometric growth dwarfs arithmetic: with a = 2, term 6 is 12 by adding and 486 by multiplying. That comparison explains compound interest better than a definition.
Task: for 3, 6, 12, 24 … find the 8th term and the sum of the first 8 (384 and 765), and state r.
Scene 3: Sum to infinity: check |r| < 1 before writing anything
With a = 8 and r = 0.5, S = a/(1 – r) = 8 / 0.5 = 16.
The formula holds only when |r| < 1. When |r| >= 1 the terms do not shrink, the partial sums diverge, and writing a total is simply wrong. So the first line of any answer is “since |r| < 1, the sum converges”.
Geometrically, halving a 8-unit square repeatedly never passes 16.
Recurring decimals are the same idea: 0.777… = 0.7/(1 – 0.1) = 7/9, which students can verify on a calculator.
The formula holds only when |r| < 1. When |r| >= 1 the terms do not shrink, the partial sums diverge, and writing a total is simply wrong. So the first line of any answer is “since |r| < 1, the sum converges”.
Geometrically, halving a 8-unit square repeatedly never passes 16.
Recurring decimals are the same idea: 0.777… = 0.7/(1 – 0.1) = 7/9, which students can verify on a calculator.
Task: write 0.454545… as a fraction (5/11) and identify a and r.
04 Question types and worked steps
| Type | Example | Steps | Answer |
|---|---|---|---|
| Calculation | General term | T(n) = a + (n – 1)d | 39 |
| Calculation | Arithmetic sum | n/2 (first + last) | 210 |
| Calculation | Geometric sum | a(r^n – 1)/(r – 1) | 728 |
| Explanation | Infinite sum | state |r| < 1 first | 16 |
Sanity check (teach this): Count terms correctly: from term 3 to term 99 there are 97 terms, not 96. The rule is last – first + 1. Test it with small numbers: 1 to 3 is 3 terms and 3 – 1 + 1 = 3. A geometric sum must exceed its last term; a sum below the last term means a sign or ratio error. An infinite sum must exceed a and stay finite.
05 Independent practice (3 questions)
| Question | Answer | What to watch |
|---|---|---|
| Arithmetic with a = 3, d = 4: find T(10) and S(10) | 39 and 210 | are they using n – 1 |
| Geometric with a = 2, r = 3: find T(6) and S(6) | 486 and 728 | sign of r – 1 |
| Write 0.454545… as a fraction | 5/11 | can they name a and r |
06 Common mistakes and what to say
| Mistake | Student writes | Say this |
|---|---|---|
| Writing T(n) = a + nd | T(10) = 3 + 40 | Substitute n = 1 and see whether you get the first term back. |
| Miscounting terms | calling 3 to 99 ninety-six terms | 99 – 3 + 1 = 97; check with 1 to 3 first. |
| Using the infinite formula without checking r | r = 2 plugged in | When r exceeds 1 the sum has no finite value, so the total does not exist. |
| Flipping the geometric sum fraction | writing (r – 1)/(a(r^n – 1)) | Numerator a(r^n – 1), denominator r – 1; swapping them wrecks the answer. |
07 Homework (paper, section and question numbers)
- Secondary Mathematics chapter index — all chapters
- DSE Maths practice paper — sequences questions 12-18
08 Exit check (last 5 minutes)
| Question | Answer | Decision |
|---|---|---|
| General term | a + (n – 1)d | correct: move to probability |
| Condition for infinite sum | |r| < 1 | wrong: reteach convergence |
| Number of multiples of 3 below 100 | 33 | two or more wrong: back to act one |
09 Tutor FAQ
Why separate r = 1 in the geometric sum?
The denominator is r – 1, which is zero at r = 1; every term is then a, so the sum is na.
When does a sum to infinity exist?
Only when |r| < 1, because the terms tend to zero and the partial sums converge to a/(1 – r).
How do I tell arithmetic from geometric?
Constant difference means arithmetic, constant ratio means geometric; in word problems look for fixed yearly additions versus fixed percentage growth.
Find a tutor — freeTutor Registration →
A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.
