Sequences & Series (AP/GP)
Interactive curriculum lessons, worked examples, and geometric problem-solving techniques designed to help Kenyan students master CBC, KPSEA, KCSE, and IGCSE mathematics.
First Principles
Objective: Distinguish between Arithmetic Progressions (AP) and Geometric Progressions (GP), determine their \(n\)-th terms, and compute partial and infinite sums.
Concrete Kenyan Scenario: Imagine two Chama investment strategies in Nairobi. In Strategy A (AP), members contribute KSh 1,000 in month 1, and increase their monthly contribution by a fixed KSh 500 every subsequent month (Month 2: 1,500, Month 3: 2,000). In Strategy B (GP), members start with KSh 1,000 and multiply their contribution by a constant factor of 1.2 (20% growth) every month. While Strategy A grows by steady addition, Strategy B accelerates exponentially.
Visual Intuition: An AP behaves like a linear staircase with constant step height \(d\). A GP behaves like compounding growth where each step is scaled by a common ratio \(r\). Explore the interactive simulator below to see how terms accumulate.
Key Formulas
1. Arithmetic Progression (AP):
\[T_n = a + (n - 1)d\] \[S_n = \frac{n}{2}\left(2a + (n - 1)d\right) = \frac{n}{2}(a + l)\]Where \(a\) is the first term, \(d = T_n - T_{n-1}\) is the common difference, \(n\) is the number of terms, and \(l\) is the last term.
2. Geometric Progression (GP):
\[T_n = a r^{n-1}\] \[S_n = \frac{a(1 - r^n)}{1 - r} = \frac{a(r^n - 1)}{r - 1} \quad (r \neq 1)\]Where \(a\) is the first term and \(r = \frac{T_n}{T_{n-1}}\) is the common ratio.
3. Sum to Infinity of a Convergent GP:
\[S_\infty = \frac{a}{1 - r} \quad \text{for } |r| < 1 \text{ (i.e., } -1 < r < 1\text{)}\]If \(|r| \ge 1\), the geometric series diverges and the sum to infinity does not exist.
Worked Examples
Problem: Find the sum of the first 5 terms of the arithmetic progression: \(3, 5, 7, 9, 11\).
- Identify the parameters: First term \(a = 3\), common difference \(d = 5 - 3 = 2\), number of terms \(n = 5\).
- Apply the AP sum formula: \[S_n = \frac{n}{2}\bigl[2a + (n - 1)d\bigr]\]
- Substitute the values: \[S_5 = \frac{5}{2}\bigl[2(3) + (5 - 1)(2)\bigr] = \frac{5}{2}[6 + 8] = \frac{5}{2}(14) = 35\]
Answer: \(35\)
Problem: The 3rd term of a GP is 45 and the 5th term is 405. Given that all terms are positive, find the 1st term \(a\) and the common ratio \(r\).
- Set up equations using \(T_n = a r^{n-1}\): \[T_3 = a r^2 = 45 \quad \text{--- (1)}\] \[T_5 = a r^4 = 405 \quad \text{--- (2)}\]
- Divide equation (2) by equation (1): \[\frac{a r^4}{a r^2} = \frac{405}{45} \implies r^2 = 9\]
- Solve for \(r\) and \(a\): Since terms are positive, \(r = 3\). Substitute \(r = 3\) into (1): \[a(3^2) = 45 \implies 9a = 45 \implies a = 5\]
Answer: \(a = 5, \; r = 3\)
Problem: A geometric progression has a first term \(a = 18\) and the sum to infinity \(S_\infty = 27\). Find the common ratio \(r\) and calculate the sum of the first 4 terms \(S_4\).
- Use the sum to infinity formula to find \(r\): \[S_\infty = \frac{a}{1 - r} \implies 27 = \frac{18}{1 - r}\] \[27(1 - r) = 18 \implies 1 - r = \frac{18}{27} = \frac{2}{3} \implies r = 1 - \frac{2}{3} = \frac{1}{3}\]
- Calculate the partial sum \(S_4\): \[S_4 = \frac{a(1 - r^4)}{1 - r} = \frac{18\left(1 - (\frac{1}{3})^4\right)}{1 - \frac{1}{3}} = \frac{18\left(1 - \frac{1}{81}\right)}{\frac{2}{3}}\] \[S_4 = 18 \times \frac{3}{2} \times \frac{80}{81} = 27 \times \frac{80}{81} = \frac{80}{3} = 26\frac{2}{3}\]
Answer: \(r = \frac{1}{3}, \; S_4 = \frac{80}{3}\)
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Practice