Lesson Notes By Weeks and Term v3 - Senior Secondary 1
Series
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Series
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Subject: Further Mathematics
Class: Senior Secondary 1
Term: 3rd Term
Week: 2
Theme: Pure Mathematics
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Define series Find the sum of series Solve problems on arithmetic and geometric progressive
An Arithmetic Progression (AP) is a sequence where the difference between consecutive terms is constant. This constant difference is called the common difference, $d$. The $n^{th}$ term of an AP is given by $T_n = a + (n-1)d$, where $a$ is the first term. The sum of the first $n$ terms of an AP, denoted as $S_n$, can be found using two main formulas: Using the first term ($a$) and common difference ($d$): $S_n = \frac{n}{2} [2a + (n-1)d]$ Using the first term ($a$) and the last term ($L$): If the $n^{th}$ term ($T_n$) is the last term ($L$), then $L = a + (n-1)d$. $S_n = \frac{n}{2} (a + L)$ Derivation (Brief explanation for teacher): Consider $S_n = a + (a+d) + (a+2d) + ... + (L-d) + L$ Writing it in reverse: $S_n = L + (L-d) + (L-2d) + ... + (a+d) + a$ Adding both equations: $2S_n = (a+L) + (a+L) + (a+L) + ... + (a+L)$ (n times) $2S_n = n(a+L)$ $S_n = \frac{n}{2} (a+L)$ Substituting $L = a + (n-1)d$ gives the first formula.
Worked Example 1 (AP Series): A civil servant, Mr. Emeka, saves ₦5,000 in his first month, ₦6,000 in his second month, ₦7,000 in his third month, and so on. a) How much will he save in his $12^{th}$ month? b) What is his total savings after 12 months?
Solution: The savings form an AP: 5000, 6000, 7000, ... First term, $a = ₦5,000$ Common difference, $d = ₦6,000 - ₦5,000 = ₦1,000$ a) Savings in the $12^{th}$ month ($T_{12}$): Using $T_n = a + (n-1)d$ $T_{12} = 5000 + (12-1)1000$ $T_{12} = 5000 + (11)1000$ $T_{12} = 5000 + 11000$ $T_{12} = ₦16,000$ Mr. Emeka will save ₦16,000 in his $12^{th}$ month. b) Total savings after 12 months ($S_{12}$): Using $S_n = \frac{n}{2} [2a + (n-1)d]$ $S_{12} = \frac{12}{2} [2(5000) + (12-1)1000]$ $S_{12} = 6 [10000 + (11)1000]$ $S_{12} = 6 [10000 + 11000]$ $S_{12} = 6 [21000]$ $S_{12} = ₦126,000$ Mr. Emeka's total savings after 12 months is ₦126,
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0. Alternatively, using $S_n = \frac{n}{2}(a + L)$: We know $a = 5000$ and $L = T_{12} = 16000$. $S_{12} = \frac{12}{2} (5000 + 16000)$ $S_{12} = 6 (21000)$ $S_{12} = ₦126,000$ A Geometric Progression (GP) is a sequence where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio, $r$. The $n^{th}$ term of a GP is given by $T_n = ar^{n-1}$, where $a$ is the first term. The sum of the first $n$ terms of a GP, denoted as $S_n$, is given by: $S_n = \frac{a(r^n - 1)}{r - 1}$, when $r > 1$ (or $r \neq 1$) $S_n = \frac{a(1 - r^n)}{1 - r}$, when $r 1$ or $r 1$, use $S_n = \frac{a(r^n - 1)}{r - 1}$ $S_5 = \frac{2000(2^5 - 1)}{2 - 1}$ $S_5 = \frac{2000(32 - 1)}{1}$ $S_5 = 2000 \times 31$ $S_5 = ₦62,000$ The total amount distributed to the first 5 women is ₦62,
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0. A GP can either be convergent or divergent. This property is determined by the common ratio ($r$).
Divergent GP: A geometric progression is divergent if the absolute value of its common ratio is greater than or equal to 1 (i.e., $|r| \ge 1$). In a divergent GP, the terms become increasingly large (or increasingly large in magnitude if negative), and therefore, the sum of its terms approaches infinity. There is no finite sum to infinity.
Examples: $r = 2$ (2, 4, 8, 16...), $r = -3$ (2, -6, 18, -54...)
Convergent GP: A geometric progression is convergent if the absolute value of its common ratio is less than 1 (i.e., $|r| 1$) is $S_n = \frac{a(r^n - 1)}{r - 1}$.
Substituting the values: $S_4 = \frac{100(1.5^4 - 1)}{1.5 - 1}$ $S_4 = \frac{100(5.0625 - 1)}{0.5}$ $S_4 = \frac{100(4.0625)}{0.5}$ $S_4 = \frac{406.25}{0.5}$ $S_4 = 812.5$ Since we cannot plant half a tree, approximately 813 trees were planted (or 812 if rounding down is preferred in context, but typically mathematical answers are given). For calculation purposes, 812.5 is the exact sum. Total trees planted in the first 4 weeks is 812.5 trees.
Commentary: Students need to correctly identify $a$, $r$, and $n$. Care must be taken with calculations involving decimals and powers. Discuss the practical implications of a non-integer number of trees. Question 3 (Convergent GP and Sum to Infinity): A pendulum swings such that the length of each successive swing is $\frac{2}{3}$ of the length of the previous swing.
If the first swing is 18 cm long: a) Write down the first three terms of the series representing the lengths of the swings. b) Is this series convergent or divergent? Give a reason. c) Calculate the total distance covered by the pendulum before it comes to rest.
Solution: This is a G
P. First term, $a = 18$ cm Common ratio, $r = \frac{2}{3}$ a)
First three terms: $T_1 = a = 18$ cm $T_2 = ar = 18 \times \frac{2}{3} = 12$ cm $T_3 = ar^2 = 18 \times (\frac{2}{3})^2 = 18 \times \frac{4}{9} = 8$ cm The first three terms are 18 cm, 12 cm, 8 cm. b) Convergent or divergent? The common ratio $r = \frac{2}{3}$. Since $|r| = |\frac{2}{3}| = \frac{2}{3}$, and $\frac{2}{3} < 1$, the series is convergent.
Reason: The absolute value of the common ratio is less than 1. c)
Total distance covered (Sum to infinity): Since the series is convergent, we can find the sum to infinity using $S_\infty = \frac{a}{1 - r}$. $S_\infty = \frac{18}{1 - \frac{2}{3}}$ $S_\infty = \frac{18}{\frac{1}{3}}$ $S_\infty = 18 \times 3$ $S_\infty = 54$ cm The total distance covered by the pendulum before coming to rest is 54 cm.
Commentary: This question tests understanding of series terms, convergence conditions, and the sum to infinity formula. Students must clearly state the reason for convergence based on the common ratio. --- Remediation Strategies (for struggling learners): Focus on Basics: Ensure mastery of sequences (identifying $a, d, r$) before moving to series. Provide ample practice on finding $T_n$ for both AP and G
P. Simplified
Examples: Start with very small values of $n$ (e.g., sum of first 3 terms) for AP and GP, allowing for manual addition to verify formula results.
Visual Aids: Use number lines for AP to visualize the common difference. For GP, use diagrams to show how terms decrease rapidly (for convergent) or increase rapidly (for divergent).
Formula Card: Provide a reference sheet with clearly labelled formulas for $T_n$ and $S_n$ for AP and GP, and $S_\infty$ for GP, along with conditions for use.
Peer Tutoring: Pair struggling learners with stronger students for one-on-one support during practice sessions.
Step-by-Step Guidance: Break down complex problems into smaller, manageable steps (e.g., "First, identify $a, d, n$. Second, write down the formula. Third, substitute values. Fourth, calculate."). Differentiation Strategies (for diverse learners): Varying Problem Complexity: Offer a range of problems from straightforward identification and calculation to multi-step word problems that require deeper analysis and interpretation.
Small Group Work: Group students heterogeneously to encourage peer learning and allow stronger students to explain concepts to those struggling.
Role-Playing: Use scenarios where students act as financial advisors or engineers solving problems related to series. Extension/Enrichment Activities (for high-achieving learners): Sigma Notation Exploration: Introduce the concept of sigma notation ($\sum$) in more depth, and ask students to write given series in sigma notation and vice versa. Have them find sums of series expressed this way.
Proof of Sum Formulas: Challenge students to derive the sum formulas for AP and GP on their own, or to explain the derivations in detail to the class.
Recurring Decimals: Explain how convergent geometric series can be used to express recurring decimals as fractions (e.g., $0.333... = 0.3 + 0.03 + 0.003 + ...$, which is a GP with $a=0.3$ and $r=0.1$).
Advanced Word Problems: Provide more intricate real-world problems that might combine AP and GP concepts, or require solving for unknown variables within the formulas (e.g., finding $n$ or $a$ given $S_n$ and other parameters).
Introduction to Other Series: Briefly introduce other types of series if time permits and student interest is high (e.g., harmonic series, binomial series, though this might be beyond SS1 scope, it can be a glimpse into higher-level maths). Focus on the idea that not all series have simple sum formulas.
Community Savings Schemes (Esusu/Ajo): In many Nigerian communities, people participate in traditional savings groups. A classic "Esusu" or "Ajo" involves fixed regular contributions.
However, some groups might have progressive contributions (e.g., each member pays a bit more in successive rounds). If contributions follow an AP, students can calculate the total amount collected by the group over a period, demonstrating the power of collective savings.
Example:* A thrift collector starts with ₦500 from a member in the first week, and increases the weekly contribution by ₦
5
0. What is the total contribution after 20 weeks? (AP sum)
Population Growth/Decay Modelling: Demographers often use geometric progression models to estimate population changes. While simplified, this concept helps understand how populations can grow exponentially (divergent GP for short term) or decline (convergent GP for rates less than 1).
Example: If a certain town's population increases by 2% each year, and its current population is 50,000, what would be the total increase in population over 10 years? (Requires calculating population each year and summing, which is a GP series for the total population, or AP for the increase if framed differently).
A simpler approach for series could be: If a disease reduces the population of a specific animal species by 1/3 each month, and there were initially 1000 animals, what is the total eventual reduction if the trend continues indefinitely? (Sum to infinity of the lost population per month). Depreciation of Assets and Investment Returns: Businesses in Nigeria purchase assets (vehicles, machinery) that depreciate in value over time. If the depreciation follows a constant percentage rate (e.g., 10% per year), the remaining value forms a GP. Conversely, investments that yield compound interest can be understood using GP, as the principal grows by a fixed ratio each period. Students can calculate the total depreciation over several years or the cumulative value of an investment.
Example:* A farmer buys a tractor for ₦4,000,
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0. It depreciates at a rate of 15% per annum. What is the total value lost due to depreciation over the first 4 years? (The amounts lost each year form a GP: $a=4,000,000 \times 0.15$, $r = 1 - 0.15 = 0.85$, then sum the depreciation amounts). ---