Grüne Brücke Lernressource für Sequences And Series

Übersicht

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.

Ziele

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)

Lektionshinweis

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.

Unterrichtsbewertung

Herzlichen Glückwunsch zum Abschluss der Lektion über Sequences And Series. Jetzt, da Sie die wichtigsten Konzepte und Ideen erkundet haben,

Sie werden auf eine Mischung verschiedener Fragetypen stoßen, darunter Multiple-Choice-Fragen, Kurzantwortfragen und Aufsatzfragen. Jede Frage ist sorgfältig ausgearbeitet, um verschiedene Aspekte Ihres Wissens und Ihrer kritischen Denkfähigkeiten zu bewerten.

Nutzen Sie diesen Bewertungsteil als Gelegenheit, Ihr Verständnis des Themas zu festigen und Bereiche zu identifizieren, in denen Sie möglicherweise zusätzlichen Lernbedarf haben.

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

Empfohlene Bücher

	Further Mathematics Untertitel Sequences and Series Verleger Mathematics Publishing House Jahr 2022 ISBN 978-1-2345-6789-0
	Mastering Arithmetic Progressions Untertitel Formulas and Applications Verleger Progression Publications Jahr 2021 ISBN 978-0-9876-5432-1

Frühere Fragen

Fragen Sie sich, wie frühere Prüfungsfragen zu diesem Thema aussehen? Hier sind n Fragen zu Sequences And Series aus den vergangenen Jahren.

WAEC SSCE - Further Mathematics - 2022 (Klicken Sie hier, um die vollständige Bewertung durchzuführen.)

Frage 1 Bericht

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.

Antwort

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

Antwortdetails