# Sequences and Series - Mathematics Form 3 Notes

## 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, Wednesday, …, Friday 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 43 = 64, 53 = 125 3, 7, 1 1 , 1 5 - , - , every term is 4 more than the previous one. To get the next term add 415 + 4 = 19, 19 + 4 =23 1/2, 2/4, 3/8 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 4/16, 5/32 ,

Example

For the nth 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 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 the differences are not constant,then it is not an arithmetic sequence

Rule for an arithmetic sequence
• The nth term of an arithmetic sequence with first term a1 and common difference d is given by:
an=a1 + (n − 1 )d

 Example Illustrations Find the general formula for the nth term, and then find the 18th term (a18) of the sequence: 4, 7, 10, 13,... We can see that the second term − the first = the third term − the second = 3, so this is the common difference. We also see that the first term is 4, so now we can plug these numbers into the general formula of the equation: an = a1+(n−1)d, or an =4+(n−1)3=4+3n−3 = 3n+1. So the general formula is an = 3n+1. To get the 18th term, we'll plug in 18 for n, so we have a18=3(18) +1=55. Find the general formula for the nth term, and then find the 12th term (a12) of the sequence: 8,−1,−10, −19,... We can see that the second term − the first = the third term − the second −1−8 = −9 (watch the negative signs!), so this is the common difference. We also see that the first term is 8, so now we can plug these numbers into the general formula of the equation: an = a1 +(n−1)d, or an = 8+(n−1)(−9) = 8−9n+9=−9n+17. So the general formula is an = −9n+17. To get the 12th term, we'll plug in 12 for n, so we have a12 =−9(12) +17 =−91. Find the general formula for the nth term, and then find the 8th term (a8) of the sequence: r, r−s, r−2s... We can see that the second term − the first = the third term − the second (r − s) − r = 0 − s = −s (watch the negative signs!), so this is the common difference. We also see that the first term is r, so now we can plug these numbers into the general formula of the equation: an = a1 +(n−1)d, or an = r+ (n−1)(−s) (we can just leave it like this). To get the 8th term, we'll plug in 8 forn, so we have a8=r−7s.

Example

Write a rule for the nth term of the sequence 50, 44, 38, 32, . . . . Then find a20.

Solution

The sequence is arithmetic with first term a1 = 50 and common difference
d = 44 − 50 = −6.

So, a rule for the nth term is:
an= a1+ (n − 1)d ..............Write general rule.
= 50 + (n − 1 )(−6) ...............Substitute for a1 and d.
= 56 − 6n ..............Simplified form.
The 20th term is a20 = 56 − 6(20) = −64.

Example

The 20th term of arithmetic sequence is 60 and the 1 6 th term is 20.Find the first term and the common difference.

Solution

a + (20-1)d = 60
a + 19d = 60………….. (1)
a +
(16-1)d = 20
a +
(15)d = 20…………..(2)
equation (1) – equation(2) gives
4d = 40
d =  0
but a + 15 d = 20
Therefore a + 15 x 1 0 =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
n − 1 = 19
n = 20

## 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
sn =n/2[2a+(n−1)d]

Note;
• If the first term (a) and the last term L are given , then
sn =n/2[a+L]

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

s8 =8/2[2a+(8−1)d]
= 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 + 28d =220 …………1
8a + 16d = 136 …………2
12 d = 84
d = 7
Substituting d =7 in equation 2 gives a = 3
Therefore,
s6 =6/2[2a+(6−1)7]
= 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:
an =a1n−1

Example

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

Solution

The first term, a =4
The common ratio, r =3
Therefore the 4
th term = 4 x 34−1
= 4 x 33
= 108
The 5
th term = 5 x 34−1
= 5 x 33
= 324
The nth term =4 x 3n−1

Example

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

1. The common ratio
2. The seventh term

Solution

The common ratio
The first term, a = 2
The 4
th term is 2 x r4−1 = 16
Thus, 2r3 = 16
r
3 = 8 (divided both sides by 2)
r = 2 (make r the subjet by dividing both sides by2)
The common ratio is 2
The seventh term =
ar6 = 2 x 26 = 128

## Geometric Series/Geometric Progression (G.P)

• The series obtained by the adding the terms of geometric sequence is called geometric series or geometric progression G.P
• The sum Sof the first n terms of a geometric series with common ratio r > 1 is:
Sn = a(rn 1)
r−1
• The sum Sn of the first n terms of a geometric series with common ratio r < 1 is:
Sna(1 − rn)
1 − r

Example

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

Solution

a = 8 , r = 24/8 = 3
Sn
a(39 1)
3 − 1
=8 (19683 − 1)
2
= 78728

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

s3 = 26 , r = 3 n = 3

26 = a(33− 1)
3−1
a(27 − 1)
2
a= 26 x 2 = 2
26
S6 =
2(36− 1)
2
= (2 x 728) = 728
2

## 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.
1. How much did he save in the second year?
2. How much did he save in the third year?
3. Find the common ratio between the savings in two consecutive years
4. How many years did he take to save the savings a sum of Ksh 58,000?
5. 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
2. If the first term is larger than the second term, find the value of r.
4.
1. 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
2. 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:
1. Last amount deposited by Amina
2. 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
1. Find the first term and common difference of the progression
2. 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
1. Write down the first four terms of the sequence
2. Find sthe sum of the fifty term of the sequence
3. 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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