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Aperçu

Welcome to the course material for 'Sequences and Series' in Further Mathematics. In this topic, we delve into the intriguing world of sequences and series, fundamental concepts that form the basis of many mathematical applications. Our primary objective is to understand the concept of sequences and series and how they are used in solving various mathematical problems.

Sequences are ordered lists of numbers that follow a specific pattern or rule. One common type of sequence is the arithmetic progression (AP), where each term is obtained by adding a constant difference to the previous term. Understanding the formula for the nth term of an AP, given by Un = U1 + (n-1)d, is crucial in identifying and working with APs effectively.

On the other hand, geometric progressions (GP) are sequences where each term is obtained by multiplying the previous term by a constant ratio. The formula for the nth term of a GP, Un = U1 * r^(n-1), is essential in recognizing and manipulating GP patterns.

Calculating the sum of finite arithmetic and geometric series is another vital aspect of this topic. For arithmetic series, we use the formula Sn = n/2 * (2a + (n-1)d), where a is the first term and d is the common difference. Similarly, the formula for the sum of a geometric series, Sn = a(1 - r^n)/(1 - r), is used to find the total sum of a geometric sequence up to the nth term.

Recurrence series, where each term is defined based on one or more previous terms, add another layer of complexity to sequences and series. Analyzing recurrence series often involves deriving explicit formulas for terms or finding patterns to predict future terms.

Understanding these concepts and formulas equips us with powerful tools to solve real-world problems that involve patterns, growth, and cumulative totals. By the end of this course material, you will be proficient in identifying, analyzing, and manipulating various types of sequences and series, paving the way for advanced studies in mathematics and its applications.

Objectifs

Calculate the sum of finite arithmetic and geometric series
Recognize geometric progressions (GP)
Understand the concept of sequences and series
Apply formulas for the nth term of an AP and GP
Analyze recurrence series
Identify and work with arithmetic progressions (AP)

Note de cours

Sequences and series form a foundational concept in mathematics and are widely used in various branches including algebra, calculus, and even in understanding complex real-life problems. This topic deals with understanding the set of ordered numbers, their properties, and the sum of these ordered numbers under specific conditions.

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Évaluation de la leçon

Félicitations, vous avez terminé la leçon sur Sequences And Series. Maintenant que vous avez exploré le concepts et idées clés, il est temps de mettre vos connaissances à lépreuve. Cette section propose une variété de pratiques des questions conçues pour renforcer votre compréhension et vous aider à évaluer votre compréhension de la matière.

Vous rencontrerez un mélange de types de questions, y compris des questions à choix multiple, des questions à réponse courte et des questions de rédaction. Chaque question est soigneusement conçue pour évaluer différents aspects de vos connaissances et de vos compétences en pensée critique.

Utilisez cette section d'évaluation comme une occasion de renforcer votre compréhension du sujet et d'identifier les domaines où vous pourriez avoir besoin d'étudier davantage. Ne soyez pas découragé par les défis que vous rencontrez ; considérez-les plutôt comme des opportunités de croissance et d'amélioration.

Find the 8th term of the arithmetic progression: 3, 6, 9, 12, ... A. 21 B. 22 C. 23 D. 24 Answer: B. 22
Calculate the sum of the first 10 terms of the geometric progression: 5, 10, 20, 40, ... A. 12,345 B. 15,639 C. 19,685 D. 24,620 Answer: A. 12,345
Identify the common ratio of the geometric progression: 2, 4, 8, 16, ... A. 2 B. 3 C. 4 D. 5 Answer: A. 2
Determine the nth term of the arithmetic progression: 1, 5, 9, 13, ... A. n + 4 B. 4n - 3 C. 4n + 1 D. 5n - 4 Answer: B. 4n - 3
Calculate the sum of the first 6 terms of the geometric progression: 3, 9, 27, 81, ... A. 1,161 B. 2,187 C. 4,374 D. 8,748 Answer: D. 8,748
Identify the common difference of the arithmetic progression: 11, 17, 23, 29, ... A. 5 B. 6 C. 7 D. 8 Answer: B. 6
Determine the 10th term of the arithmetic progression: 4, 8, 12, 16, ... A. 36 B. 38 C. 40 D. 42 Answer: C. 40
Calculate the sum of the first 5 terms of the geometric progression: 2, 6, 18, 54, ... A. 313 B. 560 C. 728 D. 972 Answer: D. 972
Identify the common ratio of the geometric progression: 3, 9, 27, 81, ... A. 2 B. 3 C. 4 D. 5 Answer: B. 3
Find the 12th term of the arithmetic progression: 7, 14, 21, 28, ... A. 76 B. 79 C. 82 D. 85 Answer: C. 82

Livres recommandés

	Further Mathematics Sous-titre Sequences and Series Éditeur Mathematics Publishing House Année 2022 ISBN 978-1-2345-6789-0
	Mastering Arithmetic Progressions Sous-titre Formulas and Applications Éditeur Progression Publications Année 2021 ISBN 978-0-9876-5432-1

Questions précédentes

Vous vous demandez à quoi ressemblent les questions passées sur ce sujet ? Voici plusieurs questions sur Sequences And Series des années précédentes.

WAEC SSCE - Further Mathematics - 2022 (Cliquez pour passer l'évaluation complète)

Question 1 Rapport

Given that nC $_{4}$ , nC $_{5}$ and nC $_{6}$ are the terms of a linear sequence (A.P), find the :

i. value of n

ii. common differences of the sequence.

Réponse

nC $_{4}$ = $\frac{n!}{[n - 4]! 4!}$ $\frac{n [n - 1] [n - 2] [n - 3] [n - 4]!}{[n - 4]! 4!}$

nC $_{5}$ = $\frac{n!}{[n - 5]! 5!}$ $\frac{n [n - 1] [n - 2] [n - 3] [n - 4] [n - 5]!}{[n - 5]! 5!}$

and nC $_{6}$ = $\frac{n!}{[n - 6]! 6!}$ → $\frac{n [n - 1] [n - 2] [n - 3] [n - 4] [n - 5] [n - 6]!}{[n - 6]! 6!]}$

d = U2 - U1 = U3 - U2

nC $_{5}$ - nC $_{4}$ = $\frac{n (n - 1) (n - 2) (n - 3) (n - 9)}{5!}$

nC $_{6}$ - nC $_{5}$ = $\frac{n (n - 1) (n - 2) (n - 3) (n - 4) (n - 11)}{5! 6}$

nC $_{5}$ - nC $_{4}$ = nC $_{6}$ - nC $_{5}$

$\frac{n (n - 1) (n - 2) (n - 3) (n - 9)}{5!}$ = $\frac{n (n - 1) (n - 2) (n - 3) (n - 4) (n - 11)}{5! 6}$

divide both sides by n(n-1)(n-2)(n-3)

$\frac{n - 9}{5!}$ = $\frac{[n - 4] [n - 11]}{5! 6}$

multiply both sides by 5! * 6

6(n-9) = (n-4)(n-11)
6n - 54 = n $^{2}$ -11n - 4n + 44
6n - 54 = n $^{2}$ - 15n + 44
n $^{2}$ - 21n + 98 = 0
n $^{2}$ - 7n - 14n + 98 = 0
n(n - 7) -14(n - 7) = 0
(n-7)(n-14) = 0
n = 7 or 14

ii.

d = U2 - U1
when n = 7
d = 7C $_{5}$ - 7C $_{4}$

d = $\frac{7 * 6 * 5!}{2! * 5!}$ - $\frac{7 * 6 * 5 * 4!}{3! * 4!}$

d = 21 - 35
d = -14

when n = 14

d = 14C $_{5}$ - 14C $_{4}$

d = $\frac{14 * 13 * 12 * 11 * 10 * 9!}{9! * 5!}$ - $\frac{14 * 13 * 12 * 11 * 10!}{10! * 4!}$

d = 2002 - 1001
d = 1001

Détails de la réponse