Sequences And Series

Gbogbo ọrọ náà

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.

Ebumnobi

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)

Akọmọ Ojú-ẹkọ

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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Ekele diri gi maka imecha ihe karịrị na Sequences And Series. Ugbu a na ị na-enyochakwa isi echiche na echiche ndị dị mkpa, ọ bụ oge iji nwalee ihe ị ma. Ngwa a na-enye ụdị ajụjụ ọmụmụ dị iche iche emebere iji kwado nghọta gị wee nyere gị aka ịmata otú ị ghọtara ihe ndị a kụziri.

Ị ga-ahụ ngwakọta nke ụdị ajụjụ dị iche iche, gụnyere ajụjụ chọrọ ịhọrọ otu n’ime ọtụtụ azịza, ajụjụ chọrọ mkpirisi azịza, na ajụjụ ede ede. A na-arụpụta ajụjụ ọ bụla nke ọma iji nwalee akụkụ dị iche iche nke ihe ọmụma gị na nkà nke ịtụgharị uche.

Jiri akụkụ a nke nyocha ka ohere iji kụziere ihe ị matara banyere isiokwu ahụ ma chọpụta ebe ọ bụla ị nwere ike ịchọ ọmụmụ ihe ọzọ. Ekwela ka nsogbu ọ bụla ị na-eche ihu mee ka ị daa mba; kama, lee ha anya dị ka ohere maka ịzụlite onwe gị na imeziwanye.

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

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Àwọn Ìwé Tó Dáa Láti Kà

Further Mathematics

Ìkọ̀wé-àbáojú: Sequences and Series

Onkọwe atẹjade: Mathematics Publishing House

Afọ: 2022

ISBN: 978-1-2345-6789-0

Mastering Arithmetic Progressions

Ìkọ̀wé-àbáojú: Formulas and Applications

Onkọwe atẹjade: Progression Publications

Afọ: 2021

ISBN: 978-0-9876-5432-1

Àwọn Ìbéèrè Tó Ti Kọjá

Nna, you dey wonder how past questions for this topic be? Here be some questions about Sequences And Series from previous years.

WAEC SSCE - Further Mathematics - 2022

Ajụjụ 1 Ripọtì

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.

Azịza

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

Akọwa Nkọwa

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

Wo Idahun

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Sequences And Series

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