Sequence and Series is one of the highest-scoring topics in ECZ Grade 12 Mathematics (Paper 1 & Paper 2). Exam candidates often struggle not because the formulas are difficult, but because they confuse Arithmetic Progressions (AP) with Geometric Progressions (GP). This definitive tutorial breaks down formulas, solving methods, and exact past exam questions step-by-step.
Note for Exam Candidates:
In ECZ Mathematics Paper 1, sequence questions are worth 3 to 4 marks and do not allow calculators. In Paper 2, structured questions carry 6 to 8 marks. Mastering both AP and GP guarantees easy marks!
1. Arithmetic Progression (AP) Fundamentals
An Arithmetic Progression (AP) is a sequence of numbers in which the difference between consecutive terms is constant. This constant difference is known as the common difference ($d$).
Key AP Formulas:
- Common Difference ($d$): $d = T_2 - T_1 = T_3 - T_2 = T_n - T_{n-1}$
- $n$-th Term ($T_n$): $T_n = a + (n - 1)d$ (where $a$ is the first term)
- Sum of First $n$ Terms ($S_n$):
Formula 1: $S_n = \frac{n}{2} [2a + (n - 1)d]$
Formula 2: $S_n = \frac{n}{2} (a + l)$ (where $l$ is the last term $T_n$)
Worked Example 1 (ECZ Paper 1 Style):
Given the AP: $5, 9, 13, 17, \dots$
(a) Find the 15th term ($T_{15}$).
Solution: First term $a = 5$. Common difference $d = 9 - 5 = 4$.
Using $T_n = a + (n-1)d$ for $n = 15$:
$$T_{15} = 5 + (15 - 1)(4) = 5 + (14 \times 4) = 5 + 56 = 61$$
(b) Find the sum of the first 20 terms ($S_{20}$).
Solution: Using $S_n = \frac{n}{2}[2a + (n-1)d]$:
$$S_{20} = \frac{20}{2}[2(5) + (20-1)(4)] = 10[10 + 19(4)] = 10[10 + 76] = 10 \times 86 = 860$$
2. Geometric Progression (GP) Fundamentals
A Geometric Progression (GP) is a sequence of numbers where each term after the first is found by multiplying the previous term by a fixed non-zero number called the common ratio ($r$).
Key GP Formulas:
- Common Ratio ($r$): $r = \frac{T_2}{T_1} = \frac{T_3}{T_2} = \frac{T_n}{T_{n-1}}$
- $n$-th Term ($T_n$): $T_n = a r^{n-1}$
- Sum of First $n$ Terms ($S_n$):
When $r > 1$: $S_n = \frac{a(r^n - 1)}{r - 1}$
When $r < 1$: $S_n = \frac{a(1 - r^n)}{1 - r}$ - Sum to Infinity ($S_\infty$): $S_\infty = \frac{a}{1 - r}$ (Only valid when $|r| < 1$, i.e., $-1 < r < 1$)
Worked Example 2 (ECZ Paper 2 Style):
The first three terms of a GP are $x + 2$, $x$, and $x - 3$.
(a) Find the value of $x$.
Solution: Since the sequence is a GP, the common ratio $r = \frac{T_2}{T_1} = \frac{T_3}{T_2}$.
$$\frac{x}{x + 2} = \frac{x - 3}{x}$$
Cross-multiplying gives: $x^2 = (x + 2)(x - 3)$
$$x^2 = x^2 - x - 6 \implies 0 = -x - 6 \implies x = -6$$
(b) Find the first term ($a$) and common ratio ($r$).
$$a = T_1 = x + 2 = -6 + 2 = -4$$
$$r = \frac{x}{x+2} = \frac{-6}{-4} = \frac{3}{2} = 1.5$$
Teacher Strategy Tip:
Encourage students to test whether a given sequence is AP or GP by calculating $T_2 - T_1$ and $\frac{T_2}{T_1}$. If $T_2 - T_1 = T_3 - T_2$, it is an AP. If $\frac{T_2}{T_1} = \frac{T_3}{T_2}$, it is a GP.
3. Sum to Infinity ($S_\infty$) Demystified
ECZ exam questions frequently ask students to calculate the sum to infinity or state the condition for $S_\infty$ to exist.
Condition for $S_\infty$: The series must be a convergent GP where the magnitude of the common ratio is strictly less than 1 ($-1 < r < 1$). If $r \ge 1$ or $r \le -1$, the series diverges and $S_\infty$ does not exist.
Worked Example 3 (Sum to Infinity):
Given the GP: $16, 8, 4, 2, \dots$
Here $a = 16$ and common ratio $r = \frac{8}{16} = 0.5$.
Since $|0.5| < 1$, sum to infinity exists:
$$S_\infty = \frac{a}{1 - r} = \frac{16}{1 - 0.5} = \frac{16}{0.5} = 32$$
Offline Revision Technique:
Copy the top 10 formula variations (finding $n$, finding $a$ and $d$ from simultaneous equations) onto flashcards for daily quick revision without needing internet access.
4. Common Pitfalls to Avoid in ECZ Exams
- Wrong Formula Selection: Using AP sum formula $S_n = \frac{n}{2}[2a+(n-1)d]$ on a Geometric Progression.
- Off-by-one Errors: Forgetting that the power in $T_n = a r^{n-1}$ is $(n-1)$, not $n$. For example, the 5th term is $a r^4$, not $a r^5$.
- Sign Errors with Negative Ratios: When $r$ is negative (e.g., $r = -\frac{1}{2}$), ensure powers are expanded carefully: $(-0.5)^3 = -0.125$ while $(-0.5)^4 = +0.0625$.
- Forgetting to State Conditions: Failing to write $-1 < r < 1$ when asked "State the condition under which the sum to infinity exists."