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Specific Objectives

By the end of the topic the learner should be able to:

(a) Identify simple number patterns;

(b) Define a sequence;

(c) Identify the pattern for a given set of numbers and deduce the general rule;

(d) Determine a term in a sequence;

(e) Recognize arithmetic and geometric sequences;

(f) Define a series

(g) Recognize arithmetic and geometric series (Progression);

(h) Derive the formula for partial sum of an arithmetic and geometric series(Progression);

(i) Apply A.P and G.P to solve problems in real life situations.

Content

(a) Simple number patterns

(b) Sequences

(c) Arithmetic sequence

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(d) Geometric sequence

(e) Determining a term in a sequence

(f) Arithmetic progression (A.P)

(g) Geometric progression (G.P)

(h) Sum of an A.P

(i) Sum of a G.P (exclude sum to infinity)

(j) Application of A.P and G.P to real life situations.

Introduction

Sequences and Series are basically just numbers or expressions in a row that make up some sort of a pattern for example,  Monday, Tuesday, Wenesday,, Friaday is a sequence that represents the days of the week. Each of these numbers or expressions are called terms or elements of the sequence.

Sequences are the list of these items, separated by commas, and series are the sum of the terms of a sequence.

Example

Sequence Next two terms

1, 8, 27, – , – Every term is cubed .The next two terms are

3, 7, 11, 15 – , – , every term is 4 more than the previous one. To get the next term add 4

15 + 4 = 19, 19 +4 =23

On the numerator, the next term is 1 more than the previous one, and the denominator, the next term is multiplied by 2 the next two terms are

Example

For the term of a sequence is given by 2n + 3, Find the first, fifth, twelfth terms

Solution

First term, n = 1 substituting (2 x 1 +3 =5)

Fifth term, n = 5 substituting (2 x 5 +3 =13)

Twelfth term, n = 12 substituting (2 x 12 +3 =27)

 

 

 

 

 

Arithmetic and geometric sequence

Arithmetic sequence.

Any sequence of a number with common difference is called arithmetic sequence

To decide whether a sequence is arithmetic, find the differences of consecutive terms. If each differences are not constant,the it is arithmetic sequence

Rule for an arithmetic sequence

The nth term of an arithmetic sequence with first term and common difference d is given by:

= + (n – 1)d

 

 

Example

Illustrations

 

Image From EcoleBooks.com

 

Example

Write a rule for the nth term of the sequence 50, 44, 38, 32, . . . . Then find .

Solution

The sequence is arithmetic with first term = 50 and common difference

d = 44 – 50 = -6. So, a rule for the nth term is:

 

= + (n – 1)d Write general rule.

= 50 + (n – 1)(-6) Substitute for a1 and d.

= 56 – 6n Simplify.

The 20th term is = 56 – 6(20) = -64.

 

Example

The 20 th term of arithmetic sequence is 60 and the 16 th term is 20.Find the first term and the common difference.

 

 

Solution

 

 

 

 

  • – (2) gives

    4d = 40

    d= 10

 

 

Therefore a + 15 x 10 =20

a + 150 = 20

a = -130

Hence, the first term is – 130 and the common difference is 10.

Example

Find the number of terms in the sequence – 3 , 0 , 3 …54

Solution

The n th term is a + ( n – 1)d

a = -30 , d =3

n th term = 54

therefore – 3 + ( n – 1) = 54

3 (n – 1 ) = 57

 

 

 

Arithmetic series/ Arithmetic progression A.P

The sum of the terms of a sequence is called a series. If the terms of sequence are 1, 2, 3, 4, 5, when written with addition sign we get arithmetic series

1 + 2 + 3 + 4 + 5

The general formulae for finding the sum of the terms is

 

Note;

If th first term (a) and the last term l are given , then

 

 

Example

The sum of the first eight terms of an arithmetic Progression is 220.If the third term is 17, find the sum of the first six terms

Solution

 

= 4( 2a + 7d )

So , 8a + 28d = 220…………………….1

The third term is a + (3 – 1)d = a + 2d =17 …………….2

Solving 1 and 2 simultaneously;

8a + 28 d =220 …………1

8a + 16 d = 136 …………2

12 d = 84

 

Substituting d =7 in equation 2 gives a = 3

Therefore,

= 3(6 x 35)

= 3 x 41

= 123

 

Geometric sequence

It is a sequence with a common ratio.The ratio of any term to the previous term must be constant.

Rule for Geometric sequence is;

The nth term of a geometric sequence with first term a1 and common ratio r is given by:

 

 

 

Example

Given the geometric sequence 4 , 12 ,36 ……find the 4th , 5th and the n th terms

Solution

The first term , a =4

The common ratio , r =3

Therefore the 4th term = 4 x

= 4 x

= 108

The 5th term = 5 x

= 5 x

= 324

The term =4 x

 

 

Example

The 4th term of geometric sequence is 16 . If the first term is 2 , find;

  • The common ration
  • The seventh term

Solution

The common ratio

The first term, a = 2

The 4th term is 2 x

Thus, 2

 

 

 

The common ratio is 2

The seventh term =

 

 

 

Geometric series

The series obtained by the adding the terms of geometric sequence is called geometric series or geometric progression G.P

The sum of the first n terms of a geometric series with common ratio r > 1 is:

 

The sum of the first n terms of a geometric series with common ratio r < 1 is:

 

Example

Find the sum of the first 9 terms of G.P. 8 + 24 + 72 +…

Solution

 

 

 

 

Example

The sum of the first three terms of a geometric series is 26 .If the common ratio is 3 , find the sum of the first six terms.

Solution

 

 

 

=

a =

 

 

 

 

End of topic  

Did you understand everything?

If not ask a teacher, friends or anybody and make sure you understand before going to sleep!

 

Past KCSE Questions on the topic.

 

 

 

1.  The first, the third and the seventh terms of an increasing arithmetic progression are three consecutive terms of a geometric progression. In the first term of the arithmetic progression is 10 find the common difference of the arithmetic progression?

2.  Kubai saved Ksh 2,000 during the first year of employment. In each subsequent year, he saved 15% more than the preceding year until he retired.

(a) How much did he save in the second year?

(b) How much did he save in the third year?

(c) Find the common ratio between the savings in two consecutive years

  1. How many years did he take to save the savings a sum of Ksh 58,000?

(e) How much had he saved after 20 years of service?

3.  In geometric progression, the first term is a and the common ratio is r. The sum of the first two terms is 12 and the third term is 16.

  1. Determine the ratio ar2

    a + ar

  (b) If the first term is larger than the second term, find the value of r.

4.  (a)  The first term of an arithmetic progression is 4 and the last term is 20. The

Sum of the term is 252. Calculate the number of terms and the common differences of the arithmetic progression

(b)  An Experimental culture has an initial population of 50 bacteria. The population increased by 80% every 20 minutes. Determine the time it will take to have a population of 1.2 million bacteria.

5.  Each month, for 40 months, Amina deposited some money in a saving scheme. In the first month she deposited Kshs 500. Thereafter she increased her deposits by Kshs. 50 every month.

Calculate the:

a)  Last amount deposited by Amina

b)  Total amount Amina had saved in the 40 months.

6.  A carpenter wishes to make a ladder with 15 cross- pieces. The cross- pieces are to diminish uniformly in length from 67 cm at the bottom to 32 cm at the top.

Calculate the length in cm, of the seventh cross- piece from the bottom

 

7.  The second and fifth terms of a geometric progression are 16 and 2 respectively. Determine the common ratio and the first term.

 

8.  The eleventh term of an arithmetic progression is four times its second term. The sum of the first seven terms of the same progression is 175

 (a)  Find the first term and common difference of the progression

 (b)  Given that pth term of the progression is greater than 124, find the least

value of P

9.  The nth term of sequence is given by 2n + 3 of the sequence

 (a)  Write down the first four terms of the sequence

 (b)  Find sn the sum of the fifty term of the sequence

 (c)  Show that the sum of the first n terms of the sequence is given by

Sn = n2 + 4n

 Hence or otherwise find the largest integral value of n such that Sn <725


 




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EcoleBooks | Mathematics Form 1-4 : CHAPTER FIFTY - SEQUENCE AND SERIES

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