Sequence And Series SS1 Further Mathematics Lesson Note

Sequence & Series

A sequence is a pattern of numbers arranged in a particular order. Each of the numbers in the sequence is called a term. The terms are related to one another according to a well-defined rule.

Consider the sequence 1, 4, 7, 10, 13 …., 1 is the first term,(T1) 4 is the second term(T2), and 7 is the third term (T3).

The sum of the terms in a sequence is regarded as a series. The series of the above sequence is

1 + 4 + 7 + 10 + 13 = 35

The nth term of a Sequence

The nth term of a sequence whose rule is stated may be represented by Tnso that T1, T2, T3etc represent the first term, second term, third term … etc respectively.

Consider the sequence 5, 9, 13, 17, 21 ……..

T1 = 5 + 4(0)

T2 = 5 + 4(1)

T3 = 5 + 4 (2)

T4 = 5 + 4 (3)

Tn = 5 + 4 (n – 1)

Tn = 5 + 4n – 4       = 4n +1

when n = 30

T30 = 4(30) + 1

T30 = 121

Examples

Write down the first four terms of the sequence whose general term is given by:

(i) Tn = n+1    

            3n +2      

(ii) Tn = 5 x (1/2)n-2

Solution

1. Tn     =   n+1

            3n + 2

T1     =  1 + 1       = 2/5

            3(1) + 2     

T2     = 2 + 1        = 3/8

          3(2) +2

T3    =  3 + 1        = 4/11

            3(3) + 2  

T4    =   4 + 1        = 5/14

                      3(4) + 2

(ii)  Tn = 5 x (1/2)n-2

T1   = 5 x (1/2)1-2   = 5(1/2)-1  = 5(2-1)-1 = 5 x 2 = 10

T2   = 5 x (1/2)2-2    = 5(1/2)0  = 5 x 1 = 5

T3   = 5 x (1/2)3-2   = 5 x (1/2) = 5/2

T4   = 5 x (1/2)4-2   = 5(1/2)2   = 5/4

The sequence is 10, 5, 5/2, 5/4 ……… ……..

Arithmetic Progression (A.P) or  Linear Sequence

An arithmetic progression (A.P) is generated by adding or subtracting a constant number from a preceding term to get a term. This constant number is called the common difference designated by the letter d. The first term is designated by a.

Ex:

T1 = a  

T2 = a + d   

T3 = a + 2d

T4 =   a + 3d   

T5 = a + 4d

So for any A.P, the nth term (Tn = Un) is given by 

Tn =Un=  a + (n – 1) d.                   

where,

Tn= Un= nth term

a = first term

d = common difference

n = no of terms

Examples

1. What is the 10th term of the sequence 10, 6, 2, -4 …..

2. Find the term of the A.P 3½, 7, 10½ ….. Which is 77.
3. The first term of an A.P. is 3 and the 8th term is 31. Find the common difference.

Solution

(1.) The A.P   = 10, 6, 2, -4

a = 10, d = 6 – 10 = – 4, n = 10

Tn  = a + (n – 1) d

T10 = 10 + (10 – 1) (-4)

T10 = 10 +9(-4)   = 10 – 36

T10 = -26.

(2.) A.P = 3 ½, 7, 10½ ……………… 77

a = 3½, d = 7 – 3½ = 3½, n =?  Tn = 77

Tn = a + (n-1)d

77 = 3½ + (n-1)3½

77 = 3½ + 3½n – 3½ 

77 = 3½ n

n = 77/3½ = 77/7/2

n = 77 x 2/7 = 22

(3) a = 3, T8 = 31, d = ? n = 8

Tn = a + (n-1) d

31 = 3 + (8-1) d

31 – 3 = 7d

d = 28/7 = 4

Arithmetic Mean

If a, b, c are three consecutive terms of an A.P, then the common difference, d, equals 

b – a = c – b    = common difference.

b + b = a +c

2b = a + c

b = ½(a +c)

Examples

(i) Insert four arithmetic means between -5 and 10.

(ii) The 8th term of a linear sequence is 18 and the 12th term is 26. Find the first term, the common difference and the 20th term.

Solution

(i) Let the sequence be -5, a, b, c, d, 10.

a = -5, T6 = 10, n =6.

Tn = a + (n-1) d

10 = -5 + (6 – 1) d

15 = 5d

d = 15/5 = 3

a = -5 + 3 = -2

b = -2 + 3 = 1

c = 1 + 3 = 4

d = 4 + 3 = 7

The numbers will be -5, -2, 1, 4, 7, 10.

(ii) T8 = a + 7d = 18, T12 = a +11d = 26

a + 7d = 18    ……………….. (i)

a + 11d =26   ……………….. (ii)

Subtract (i) from (ii)

4d = 8

d = 2

Substitute for d = 2 in (i)

a + 7 (2) = 18

a = 18 – 14

a = 4

T20 = a + (n – 1) d   = a + 19d

T20 = 4 + (20 – 1) 2

= 4 + 19 x 2

T20   = 42

The sum of Terms in an A.P

To find an expression for the sum of n terms of a linear sequence, Let Sn be the sum, and then 

Sn = a + (a + d) + (a + 2d) + ……. + Tn ………….. (i)

Also

Sn = Tn+ (Tn- d) + (Tn- 2d) + ……… a ………. (ii)

Adding (1) and (2)

2Sn = (a + Tn) + (a + Tn) + (a + Tn) + ………… + (a + Tn)

(2Sn = n (a + Tn)

(Sn = n/2 (a + Tn)

But Tn = a + (n-1) d

Sn = n/2 (2a + (n-1) d)

Examples

(1) Find the sum of the first 25 terms of the A.P 3, 10, 17 ……….

(2) Find the sum of the first eight terms of a linear sequence whose first term is 6 and the last term is 46.

(3) The sum of the first ten terms of an arithmetic progression is 255. Find the sum of the next twenty terms of the A.P. if the sum of the first twenty terms is 1010.

Solution

1. A.P = 3, 10, 17 …………..

a = 3, d = 7, n = 25      

/Sn = n/2 (2a + (n-1) d)

= 25/2 (2 x3 + (25 – 1) 7)

Sn = 25/2 (6 +24 x 7)

S25 =25/2 x174   = 2175

2. A.P , a = 6,  Tn = 46, n = 8

Sn =   n/2 (a + Tn)

= 8/2 (6 + 46)

Sn = 4 (52) = 208.

3. S10 = 10/2 (2a + (10 – 1) d) = 255

S20 = 20/2 (2a + (20 – 1) d) = 1010

  5 (2a + 9d) = 255

  10 (2a + 19d) = 1010

   2a + 9d = 51 …………….(i)

   2a +19d = 101 ……………… (ii)

        Subtract (i) from (ii) 

        10d = 50

         d = 5

         Substitute for d = 5 in (i)

2a + 9 x 5 = 51

2a = 51 – 45

2a = 6

a = 3

The sum of the next 20 terms = S30 – S10

S30 = 30/2 (2 x 3 + (30 -1) 5)

        = 15 (6 + 29 x 5)

           S30 = 2265

S30 – S10 = 2265 – 255

    = 2010

ASSIGNMENT 

1. The sum of the first four terms of a linear sequence (A.P) is 26 and that of the next four terms is 74.

Find the values of (i) the first term and (ii) the common difference.

2. Calculate the (i) common difference (ii) the 20th term of the arithmetic progression; 

     100, 96, 92, 88, 86…

3. The first three terms of an A.P are x, 2x+1, 4x+1, find x and the sum of the first 18 terms.
4. The sum of the first twenty –terms of an A.P. is 28, and the sum of the first twenty-eight terms is 21. Find which terms of the sequence is o and also the sum of the term preceding it.