# Arithmetic and Geometric Progressions

> Additional Mathematics · CIE IGCSE 0606
> Source: https://www.owlsprep.com/study/cie-0606-u12-arithmetic-and-geometric-progressions/

This guide covers core AP/GP skills for CIE IGCSE Additional Mathematics 0606, including identifying progressions, calculating nth terms, finite sums, and convergent infinite GP sums, with exam-aligned worked examples.

**Prerequisites:** Basic algebra manipulation and solving linear/quadratic equations; Absolute value and basic inequality rules

## Learning objectives

- Distinguish between arithmetic (AP) and geometric (GP) progressions from given sequences
- Calculate the nth term and sum of first n terms for APs and GPs using provided formulae
- Justify convergence of a GP and calculate its sum to infinity when |r| < 1
- Solve exam-style word problems combining AP and GP concepts

## Identifying Arithmetic vs Geometric Progressions

**Arithmetic Progression (AP)** — A sequence where consecutive terms differ by a constant common difference $d$, which can be positive, negative, or zero.

*Notation:* $a, a+d, a+2d, a+3d, ...$

*Example:* 2, 5, 8, 11, ... (d = 3)

**Geometric Progression (GP)** — A sequence where each consecutive term is the previous term multiplied by a constant common ratio $r$, which can be positive, negative, integer or fractional.

*Notation:* $a, ar, ar^2, ar^3, ...$

*Example:* 4, 2, 1, 0.5, ... (r = 1/2)

> **tip**
>
> To confirm a sequence is AP: check $u_{n+1} - u_n$ is constant for all n. To confirm GP: check $\frac{u_{n+1}}{u_n}$ is constant for all n.

**Worked example:** Classify the following sequences as AP, GP, or neither: (a) 12, 7, 2, -3, ... (b) 3, 9, 27, 81, ... (c) 1, 4, 9, 16, ...

1. For (a): Calculate difference between consecutive terms: $7-12=-5$, $2-7=-5$, $-3-2=-5$. Constant difference = -5 → AP.
2. For (b): Calculate ratio between consecutive terms: $9/3=3$, $27/9=3$, $81/27=3$. Constant ratio = 3 → GP.
3. For (c): Differences: $4-1=3$, $9-4=5$, $16-9=7$ (not constant). Ratios: $4/1=4$, $9/4=2.25$, $16/9≈1.78$ (not constant) → neither.

**Check your understanding**

1. Is the sequence 2, -4, 8, -16, ... an AP, GP, or neither?

   - AP
   - GP
   - Neither

   *Why:* Correct, the common ratio is -2, which is constant for all consecutive terms.

2. What is the common difference of the AP 10, 5, 0, -5, ...?

   - 5
   - 0
   - -5
   - 10

   *Why:* Correct, subtract consecutive terms: $5-10=-5$, which is constant.

## nth Term and Sum of First n Terms for AP

The nth term of an AP and sum of first n terms are provided in your exam formula booklet. For an AP with first term $a$, common difference $d$, and number of terms $n$:

$$u_n = a + (n-1)d, \quad S_n = \frac{n}{2}[2a + (n-1)d] = \frac{n}{2}(a + l)$$

> **note**
>
> $l$ denotes the last term of the AP, which can be used as a shortcut if you know the first and final term of the sequence you are summing.

**Worked example:** An AP has first term 7 and common difference -2. Find (a) the 15th term, (b) the sum of the first 20 terms.

1. Identify given values: $a=7$, $d=-2$. For part (a), $n=15$.
2. Substitute into nth term formula:

   $$u_{15} = 7 + (15-1)(-2) = 7 - 28 = -21$$
3. For part (b), $n=20$, substitute into sum formula:

   $$S_{20} = \frac{20}{2}[2(7) + (20-1)(-2)] = 10(14 - 38) = -240$$

> **Exam tip:** When given the sum of terms, rearrange the sum formula to form a quadratic equation if needed, a common exam question style. Always state the values of a, d, n you are using to avoid arithmetic errors.

## nth Term and Sum of First n Terms for GP

For GPs, the nth term and finite sum formulae are also provided in the exam booklet. For a GP with first term $a$, common ratio $r$, and $n$ terms:

$$u_n = ar^{n-1}, \quad S_n = \frac{a(1 - r^n)}{1 - r} \quad (r \neq 1)$$

> **tip**
>
> Use the rearranged sum formula $S_n = \frac{a(r^n - 1)}{r - 1}$ when $r>1$ to avoid negative numerators and denominators, reducing arithmetic errors.

**Worked example:** A GP has first term 2 and common ratio 3. Find (a) the 6th term, (b) the sum of the first 8 terms.

1. Identify given values: $a=2$, $r=3$. For part (a), $n=6$.
2. Substitute into nth term formula:

   $$u_6 = 2 \times 3^{6-1} = 2 \times 243 = 486$$
3. For part (b), $n=8$, use the sum formula for $r>1$:

   $$S_8 = \frac{2(3^8 - 1)}{3 - 1} = 6561 - 1 = 6560$$

**Worked example:** The 3rd, 5th and 8th terms of an arithmetic progression are three consecutive terms of a geometric progression. The common difference of the AP is not zero. Find the common ratio of the GP.

1. Step 1: Write the three AP terms using $u_n = a + (n-1)d$:

   $$u_3 = a + 2d, \quad u_5 = a + 4d, \quad u_8 = a + 7d$$
2. Step 2: Three consecutive GP terms satisfy (middle)$^2$ = (first)(last), so the middle term is the geometric mean:

   $$(a + 4d)^2 = (a + 2d)(a + 7d)$$
3. Step 3: Expand both sides:

   $$a^2 + 8ad + 16d^2 = a^2 + 9ad + 14d^2$$
4. Step 4: Cancel $a^2$ and collect terms:

   $$2d^2 = ad$$
5. Step 5: Since $d \neq 0$, divide by $d$ to get $a = 2d$. Substitute back to find the three GP terms:

   $$u_3 = 4d, \quad u_5 = 6d, \quad u_8 = 9d$$
6. Step 6: The common ratio is the ratio of consecutive terms:

   $$r = \frac{u_5}{u_3} = \frac{6d}{4d} = \frac{3}{2}$$

**Check your understanding**

1. A GP has $a=16$, $r=1/2$. Find $S_4$.

   *Why:* Correct: $S_4 = 16 \times \frac{1 - (0.5)^4}{1 - 0.5} = 30$

## Sum to Infinity of a Convergent GP

**Convergent Geometric Progression** — A GP where the absolute value of the common ratio is less than 1 ($|r| < 1$), meaning terms get progressively closer to zero, so the sum of an infinite number of terms is a fixed finite value.

*Notation:* $S_\infty = \frac{a}{1 - r}$

> **warning**
>
> You must explicitly state that $|r| < 1$ (and prove it if $r$ is not given) in your exam answer when using the sum to infinity formula to get full marks. If $|r| \geq 1$, the GP is divergent and has no finite sum to infinity.

**Worked example:** A GP has first term 18 and common ratio 1/3. (a) Justify that the GP is convergent. (b) Find its sum to infinity.

1. For part (a): Calculate $|r| = |1/3| = 1/3 < 1$, so the GP is convergent.
2. For part (b), substitute $a=18$ and $r=1/3$ into the sum to infinity formula:

   $$S_\infty = \frac{18}{1 - 1/3} = \frac{18}{2/3} = 27$$

> **Exam tip:** Logarithms may be used to solve for $n$ in GP problems where you need to find the smallest $n$ such that $u_n$ or $S_n$ meets a given condition, which is allowed per the 0606 syllabus.

## Common pitfalls

- **Wrong:** Confusing AP and GP formulae, using addition instead of multiplication for GP terms
  - Why it fails: Students mix up the structure of AP and GP sequences, leading to incorrect substitutions
  - Correct: Always confirm the progression type first, write down a, d/r explicitly before substituting into any formula
- **Wrong:** Using the sum to infinity formula for a GP with |r| ≥ 1
  - Why it fails: Forgetting the convergence condition leads to invalid finite sum values for divergent GPs, losing method marks
  - Correct: Every time you use $S_\infty$, first state and verify that |r| < 1 in your answer
- **Wrong:** Using $n$ instead of $n-1$ in the exponent for GP nth term
  - Why it fails: The first term is $ar^0$, so the nth term has exponent n-1, not n, leading to off-by-one errors
  - Correct: Test your nth term formula with n=1: you should get the given first term a
- **Wrong:** Rounding intermediate values when calculating GP terms or sums
  - Why it fails: Rounding early propagates errors, especially for non-integer values of r
  - Correct: Keep intermediate values in exact fractional or exponent form until the final step, then round as required
- **Wrong:** Assuming all sequences are either AP or GP
  - Why it fails: Many exam questions include sequences that are neither to test your ability to verify progression type
  - Correct: Always check for constant difference (AP) or constant ratio (GP) before applying any AP/GP formulae

## Cheatsheet

| Progression Type | nth Term Formula | Sum of First n Terms | Sum to Infinity (if valid) |
| --- | --- | --- | --- |
| Arithmetic Progression (AP) | $u_n = a + (n-1)d$ | $S_n = \frac{n}{2}[2a + (n-1)d] = \frac{n}{2}(a+l)$ | AP has no finite sum to infinity |
| Geometric Progression (GP) | $u_n = ar^{n-1}$ | $S_n = \frac{a(1-r^n)}{1-r}$ ($r≠1$) | $S_\infty = \frac{a}{1-r}$, only valid if $\|r\| < 1$ |

## What's next

Now that you have mastered AP and GP core skills, you are ready to tackle combined series problems and word problems commonly tested in the CIE 0606 exam. Progressions are often paired with algebra and logarithms to solve for unknown values of n, d, or r, so reinforcing your algebra manipulation skills will help you tackle more complex exam questions efficiently. In your next study sessions, practice past paper questions that mix AP and GP concepts, as these are high-value questions that test your ability to distinguish between progression types and apply the correct formulae. You can also apply your series skills to practical scenarios like compound interest and population growth problems, which are common exam contexts.

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