Arithmetic & Geometric Mastery

Arithmetic & Geometric Progressions, $S_\infty$ & Financial Series

Progressions quantify sequential growth and recurring patterns. Master Arithmetic Progressions ($a, d, T_n, S_n$), Geometric Progressions ($a, r, T_n, S_n$), infinite convergence ($S_\infty = \frac{a}{1-r}$), converting recurring decimals to fractions, and modeling loan amortizations.

1. Core Progression Formulas

Arithmetic Progression (AP)

$d = T_2 - T_1 = T_3 - T_2$
$T_n = a + (n - 1)d$
$S_n = \frac{n}{2}[2a + (n - 1)d] = \frac{n}{2}[a + l]$
  • • Crucial link: $T_n = S_n - S_{n-1}$ for $n \ge 2$.
  • • 3 consecutive terms in AP: $x - d, x, x + d$ or $2b = a + c$.

Geometric Progression (GP)

$r = \frac{T_2}{T_1} = \frac{T_3}{T_2}$
$T_n = a r^{n-1}$
$S_n = \frac{a(1 - r^n)}{1 - r} = \frac{a(r^n - 1)}{r - 1}$
$S_\infty = \frac{a}{1 - r} \quad (-1 < r < 1)$
  • • 3 consecutive terms in GP: $\frac{b}{a} = \frac{c}{b} \implies b^2 = ac$.
  • • $S_\infty$ exists only if common ratio $|r| < 1$.

2. SPM Examiner Pitfalls in Progressions

Trap 1: Confusing "Sum of terms from $T_m$ to $T_n$":

Sum from the 6th to the 15th term is $S_{15} - S_5$, NOT $S_{15} - S_6$! Subtracting $S_6$ accidentally removes the 6th term itself.

Trap 2: Sign of Inequality when Dividing by $\log r$ ($r < 1$):

When solving $ar^{n-1} < k$ or $1 - r^n > 0.99$ where $r < 1$ (e.g. $r = 0.8$), $\log(0.8) \approx -0.0969$ is a negative number! Dividing both sides by $\log(0.8)$ MUST flip the inequality sign ($\ge$ becomes $\le$).

Trap 3: Bouncing Ball Rebound Distance:

A ball dropped from height $H$ travels downward $H$ on the first drop, then rebounds up and down $2 \times rH$, $2 \times r^2H$, etc. Total distance $= H + 2 \times \frac{rH}{1-r}$, NOT simply $\frac{H}{1-r}$!

3. Progressive Worked Examples with SPM Marking Rubrics

Example 1 • AP Consecutive Terms & Sum [5 Marks]
The first three terms of an arithmetic progression are $2k + 1$, $4k - 2$, and $5k + 3$.

(a) Find the value of $k$ and state the common difference $d$. [3 marks]

(b) Calculate the sum of the first 20 terms of this progression. [2 marks]

(a) In an AP, common difference $d = T_2 - T_1 = T_3 - T_2$: [1m: P1: Equate common differences]
$(4k - 2) - (2k + 1) = (5k + 3) - (4k - 2) \implies 2k - 3 = k + 5 \implies k = 8$. [1m: N1: $k = 8$]
First term $a = 2(8) + 1 = 17$. Common difference $d = 2(8) - 3 = 13$. [1m: N1: $d = 13$]
(b) $S_{20} = \frac{20}{2}[2(17) + (20 - 1)(13)] = 10[34 + 19(13)] = 10[34 + 247]$. [1m: K1: Formula substitution]
$S_{20} = 10(281) = 2,810$. [1m: N1: Correct sum]
Example 2 • Sum to Infinity & Recurring Decimals [4 Marks]
Express the recurring decimal $0.2454545\dots$ as a fraction in its simplest form using the sum to infinity of a geometric progression.
$0.2454545\dots = 0.2 + [0.045 + 0.00045 + 0.0000045 + \dots]$. [1m: P1: Decompose into non-recurring + GP]
Inside bracket is GP with $a = 0.045 = \frac{45}{1000}$ and common ratio $r = \frac{0.00045}{0.045} = 0.01 = \frac{1}{100}$. [1m: K1: Identify $a$ and $r$]
$S_\infty = \frac{a}{1 - r} = \frac{0.045}{1 - 0.01} = \frac{0.045}{0.99} = \frac{45}{990} = \frac{1}{22}$. [1m: K1: Calculate $S_\infty$]
Total $= 0.2 + \frac{1}{22} = \frac{1}{5} + \frac{1}{22} = \frac{22 + 5}{110} = \frac{27}{110}$. [1m: N1: Fraction in simplest form]
Example 3 • KBAT Bouncing Elastic Ball Physics [6 Marks]
An elastic rubber ball is dropped from a vertical height of $15\text{ meters}$ above a flat concrete floor. Each time the ball bounces off the floor, it rebounds to a height of $\frac{3}{5}$ of the height from which it fell.

(a) Calculate the maximum height, in $\text{m}$, reached by the ball on its 4th bounce. [2 marks]

(b) Calculate the total vertical distance, in $\text{m}$, travelled by the ball before it comes to rest. [4 marks]

(a) Rebound heights form a GP with $T_1 = 15 \times \frac{3}{5} = 9\text{ m}$ and $r = \frac{3}{5} = 0.6$. [1m: K1: First rebound height $a = 9$]
4th bounce: $T_4 = a r^{4-1} = 9(0.6)^3 = 9(0.216) = 1.944\text{ meters}$. [1m: N1: $T_4 = 1.944\text{ m}$]
(b) Initial drop is only downward: $15\text{ m}$. Every rebound travels UP and DOWN ($2 \times h_n$). [1m: P1: Physical distance structure identified]
$\text{Total Distance} = 15 + 2[T_1 + T_2 + T_3 + \dots] = 15 + 2 \times S_\infty$. [1m: K1: Formulation with $2 S_\infty$]
$S_\infty = \frac{a}{1 - r} = \frac{9}{1 - 0.6} = \frac{9}{0.4} = 22.5\text{ m}$. [1m: K1: Compute $S_\infty = 22.5$]
$\text{Total Distance} = 15 + 2(22.5) = 15 + 45 = 60\text{ meters}$. [1m: N1: Total distance $= 60\text{ m}$]
Open Form 4 Chapter 5 Worksheet