Recurso de Aprendizaje Green Bridge Para Sequences And Series

Resumen

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.

Objetivos

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)

Nota de la lección

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.

Evaluación de la lección

Felicitaciones por completar la lección del Sequences And Series. Ahora que has explorado el conceptos e ideas clave, es hora de poner a prueba tus conocimientos. Esta sección ofrece una variedad de prácticas Preguntas diseñadas para reforzar su comprensión y ayudarle a evaluar su comprensión del material.

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Utiliza esta sección de evaluación como una oportunidad para reforzar tu comprensión del tema e identificar cualquier área en la que puedas necesitar un estudio adicional. No te desanimes por los desafíos que encuentres; en su lugar, míralos como oportunidades para el crecimiento y la mejora.

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

Libros Recomendados

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

Preguntas Anteriores

¿Te preguntas cómo son las preguntas anteriores sobre este tema? Aquí tienes una serie de preguntas sobre Sequences And Series de años anteriores.

WAEC SSCE - Further Mathematics - 2022 (Haz clic para realizar la evaluación completa)

Pregunta 1 Informe

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.

Respuesta

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

Detalles de la respuesta