# SEQUENCES AND SERIES

## About Course

In Grade 10 and 11 you learnt about linear and quadratic number patterns. Linear

number patterns have a constant difference between consecutive terms while

quadratic number patterns have a constant second difference.

REVISION OF QUADRATIC NUMBER PATTERNS

In Grade 11 we dealt with quadratic number patterns having a general term of the

form 2 Tn = ++ an bn c .

EXAMPLE

Consider the following number pattern: 2 ; 3 ; 6 ;11; …

(a) Determine the nth term (general term) and hence the value of the 42nd term.

(b) Determine which term will equal 1091.

Solutions

(a)

2

2

42

22 3 1 2

1 3(1) 1 (1) ( 2) 2

2 3

T 23

T (42) 2(42) 3 1683

n

a ab abc

a b

b c

n n

= += ++=

∴ = ∴ + = ∴ + − =

∴ = − ∴ =

∴ = − +

∴ = − + =

(b) T 1091 n =

2

2

2 3 1091

2 1088 0

( 34)( 32) 0

34 or 32

But 32

34

The 34th term will equal 1091

∴ − + =

∴− − =

∴ − + =

∴ = = −

≠ −

∴ =

n n

n n

n n

n n

n

n

REVISION EXERCISE

1. Determine the general term for each number pattern below.

2 3 6 11

1 3 5

2 2

abc + +

3a b +

2a

(a) 2; 6 ; 14 ; 26 ; … (b) 4 ; 9;16 ; 25 ; … (c) 1; 3 ; 6 ;10 ; …

(d) −1; 0 ; 3 ; 8 ; … (e) −−− − 3 ; 6; 11; 18 ; … (f) 10 ; 6 ; 3 ;1; …

2

2. Consider the following number pattern: 1; 6 ;15 ; 28 ; …

(a) Determine the general term and hence the value of the 20th term.

(b) Determine which term will equal 3160.

3. Consider the following number pattern: −− − − 4 ; 10 ; 18 ; 28 ; …

(a) Determine the general term and hence the value of the 25th term.

(b) Determine which term will equal −810.

4. Consider the following pattern that emerges when you add the terms of a

linear number pattern.

First the linear number pattern: 3 ; 7 ;11;15 ; …

Line 1: 3

Line 2: 3 7 10

Line 3: 3 7 11 21

Line 4: 3 7 11 15 36

+ =

++ =

++ + =

.

.

.

(a) Determine the sum of the numbers in Line 5 and 6.

(b) What type of number pattern is formed by the sum of the numbers in

each line?

(c) Hence or otherwise determine the sum of the numbers in Line n.

ARITHMETIC SEQUENCES

Consider the following linear number pattern: 7 ; 10 ; 13 ; 16 ; 19 ; …

We can rewrite this pattern using only the first term and the constant difference.

T 7 1 = and the constant or common difference = T T T T T T … 3 21 32 43 − = − = − = =

T 73 2 = +

T 7 3 3 7 2(3) 3 = ++= +

T 7 3 3 3 7 3(3) 4 = +++= +

T 7 3 3 3 3 7 4(3) 5 = ++++= +

T 7 9(3) 10 = +

T 7 99(3) 100 = +

∴T 7 1 (3) n = + (n − )

Therefore the general term of the pattern is: T 7 ( 1)(3) n = + n −

The general term can be simplified further as follows:

T 7 ( 1)(3)

T 73 3

T 43

n

n

n

n

n

n

= + −

∴ = + −

∴ = +

If the letter a is used for the first term and d for the constant difference of a linear

pattern, then the pattern can be written as follows:

T1 = a

T2 = + a d

T( ) 2 3 = + +=+ ad d a d

T ( 2) 3 4 = + +=+ a d da d

3

T ( 3) 4 5 = + +=+ a d da d

6

10

100

T 5

T 9

T 99

a d

a d

a d

= +

= +

= +

T ( 1) n ∴ = +an d −

Therefore the general term of the pattern is: T ( 1) n = + an d −

Linear patterns are also called arithmetic sequences and have a general term of

T ( 1) n = + an d − where:

EXAMPLE 1

Consider the arithmetic sequence 3 ; 5 ; 7 ; 9 ; …

(a) Determine a formula for the general term of the above sequence.

(b) Find the value of the 50th term.

Solution

(a) T 3 1 =

∴a = 3

2 1

3 2

T T 5 3 2

T T 75 2

d 2

− = − =

− = − =

∴ =

To find the general term Tn you have to substitute a = 3 and d = 2 into the

general term for an arithmetic sequence, namely T ( 1) n = + an d −

T 3 ( 1)(2)

T 32 2

T21

n

n

n

n

n

n

∴ = + −

∴ = + −

∴ = +

(b) T21 n = +n

∴T 2(50) 1 101 50 = +=

EXAMPLE 2

Consider the following sequence −− − − 5 ; 9 ; 13 ; 17 ;…

(a) Show that the sequence is arithmetic.

(b) Find the value of the 25th term of the sequence.

Solutions

(a) T T 9 ( 5) 4 2 1 − = − −− = − (b) T ( 1) n = + an d −

T T 13 ( 9) 4 2 1

There is a constant difference of 4

− = − −− = −

− 25

5, 4 and 25

T 5 (25 1)( 4) 101

ad n = − = − =

∴ = − + − − = −

a represents the first term

d represents the constant or common difference

n represents the position of the term

Tn represents the nth term or general term (the value of the term in the nth position)

4

EXAMPLE 3

Determine which term of the sequence −1; 2 ; 5 ; 8 ; … is equal to 80.

Solution

T T 2 ( 1) 3 and T T 5 2 3 21 32

d 3 (the sequence is arithmetic)

− = − − = − = − =

∴ =

Write down what is given:

1, 3 and T 8 but n 0 (we know t ot the posit he actual term’s value ion or value) n a d = − = = n

T ( 1) n = + an d − (state the formula)

∴80 1 ( 1)(3) = − + n − (substitute 1, 3 and T 80 n a d = − = = )

80 1 3 3

84 3

28

n

n

n

∴ = − + −

∴ =

∴ =

∴T 80 28 =

EXAMPLE 4

xx x ; 4 5 ;10 5 + − are the first three terms of an arithmetic sequence.

Determine the value of x and hence the sequence.

Solution

1

2

3

T 5

T 4 5 4(5) 5 25

T 10 5 10(5) 5 45

The sequence is 5 ; 25 ; 45 ; ……..

x

x

x

∴ = =

∴ = += +=

∴ = − = − =

∴

EXERCISE 1

1. Determine the general term of the following arithmetic sequences:

(a) −1;3 ; 7 ;…. (b) 4 ; 2 ; 8 ; …… − − (c) 1; 1; 3 ; …. − −

(d) 99;106;113;….

2. Determine the 38th term for each of the following arithmetic sequences:

(a) −−− 4; 8; 12;…. (b) 2; 1,5; 5;… − − (c) 99; 88; 77;….

(d) 21 9 6; ; ; …. 4 2 (e) T34 k = k − (f) T 25 k = − k +

2 1

3 2

Since the sequence is arithmetic, it is clear that

T T (4 5) ( ) 3 5

T T (10 5) (4 5) 10 5 4 5 6 10

3 5 6 10

3 15

5

d x xx

d x x xxx

x x

x

x

= − = + − = −

= − = − − + = −− − = −

∴ + = −

∴− = −

∴ =

5

3. (a) Which term of the arithmetic sequence − − 5; 2;1;…. is equal to 94?

(b) Which term of the arithmetic sequence 4 ; 2,5 ; 1; … is equal to

(c) Find the number of terms in the sequence 12 ; 7 ;2 ; ….; 203 − .

(d) Find the number of terms in the sequence −− − 55; 48; 41;….; 85

4. Given the arithmetic sequence: −− − 29; 23; 17;….

(a) Determine the 31st term.

(b) Determine which term is equal to 31.

5. Given the arithmetic sequence: − −− 13; 9; 5;….

(a) Which term in the above sequence is 51?

(b) Calculate the 51st term.

6. pp p ; 2 2 ; 5 3 ; … + + are the first three terms of an arithmetic sequence.

(a) Calculate the value of p.

(b) Determine the sequence.

(c) Find the 49th term.

(d) Which term of the sequence is 1

2 100 ?

7. xxx +++ 3 ; 2 6 ; 3 9 ; … are the first three terms of an arithmetic

sequence.

(a) Determine the 10th term in terms of x.

(b) Determine the nth term in terms of x.

GEOMETRIC SEQUENCES

Consider the following number pattern: 6 ; 12 ; 24 ; 48 ; 96 ; …

This pattern is not linear (arithmetic) since there is no constant difference between

the terms. In this pattern each successive term is obtained by multiplying the

previous term by 2.

Notice too that the following ratios between the terms are also equal to 2:

2 4 3

123

T T T … 2

TTT = = ==

There is a constant ratio for this number pattern and this kind of number pattern is

called an exponential or geometric sequence.

We can rewrite this pattern using only the first term and the constant ratio:

T 6 1 =

T 62 2 = ×

2 T 622 62 3 = × × = ×

3 T 6222 62 4 = ××× = ×

4 T 62 5 = ×

9 T 62 10 = ×

99 T 62 100 = ×

1 T 62n

n

− ∴ = ×

Therefore the general term of the pattern is: 1 T 62n

n

− = ×

If the letter a is used for the first term and r for the constant ratio of an exponential

number pattern, then the pattern can be written as follows:

−66,5?

6

T1 = a

T2 = a r ar × =

2 T( ) 3 = a r r ar × × =

2 3 T( ) 4 = a r r ar × × =

4 T5 = ar

9 T10 = ar

99 T100 = ar

1 T n

n ar − ∴ =

Therefore the general term of the pattern is: 1 T n

n ar − =

Exponential or geometric number patterns have a general term: 1 T n

n ar − =

where:

EXAMPLE 5

Consider the geometric sequence 2 ; 3 ; 4,5 ; 6, 25 ; …

(a) Determine a formula for the general term of the sequence.

(b) Find the value of the 15th term.

Solutions

(a) T 2 1 = = a

2 3

1 2

T 3 4,5 3 T and

T2 T 3 2

3

2

r

= ==

∴ =

Substitute a = 2 and 3

2

r = into the general term 1 T n

n ar − =

1 3 T 2

2

n

n

− =

(b)

1 3 T 2

2

n

n

− =

15 1

15

14

15

15

3 T 2

2

3 T 2

2

T 583,86 (rounded off to two decimal places)

− =

∴ =

∴ =

a represents the first term

d represents the constant or common ratio

n represents the position of the term

Tn represents the nth term or general term (the value of the term in the nth position)

7

EXAMPLE 6

Consider the following sequence − − 5 ;10 ; 20 ; 40 ;…

(a) Show that the above sequence is geometric.

(b) Determine the value of the 20th term of the above sequence.

Solution

(a) 2 3

1 2

T 10 20 T 2 and 2

T 5 T 10

− = = − = = − − ( constant ratio 2 = − )

(b) 1 T with 5, 2 and 20 n

n ar a r n − = = − = − =

20 1 19 T ( 5)( 2) ( 5)( 2) 2 621 440 20

− ∴ = − − = − − =

Revision of basic exponential equations and other types

Exponential equations and equations involving odd or even exponents form an

integral part of the examples and exercises which follow in this chapter. It is

therefore advisable to revise these basic equations before proceeding.

Consider the following examples of exponential equations:

Solve for n:

(a) 64 2n = (b) 64 ( 2)n = −

6 2 2

6

6

n

n

n

∴ =

∴ =

∴ =

6 ( 2) ( 2)

6

6

n

n

n

∴ − = −

∴ =

∴ =

(c) 1 1

81 3

n = (d) 1 1

81 3

n = −

4 1 1

3 3

4

4

n

n

n

∴ =

∴ =

∴ =

4 1 1

3 3

4

4

n

n

n

∴ − =

∴ =

∴ =

(d) 1 1 2

128 2

n = −

8

11 1 1 2

128 2 2 2

1 1

256 2

1 1

2 2

8

n

n

n

n

∴ × = − ×

∴ = −

∴ − = −

∴ =

8

Consider the following equations involving odd or even exponents:

(a) 2

r = 9

This equation can be solved by taking the square root on on both sides

∴r = ± 9 (Don’t forget to include ± since there are two solutions)

∴r = ±3

(b) 3 r = 27

In this equation, there is no need to include ± since there is only one

solution. Simplify take the cube root on both sides:

3 27

3

r

r

∴ =

∴ =

(c) 3 r = −27

3 27

3

r

r

∴ = −

∴ = −

(d) 4

r =16

This equation can be solved by taking the fourth root on on both sides

4 ∴r = ± 16 (Don’t forget to include ± since there are two solutions)

∴r = ±2

Summary

Consider the equation n r a =

If n is odd, then there is one solution: n r a =

If n is even, then there are two solutions: n r a = ± provided a > 0

EXAMPLE 7

Determine which term in the sequence 4 ;12 ; 36 ;108 ; … is equal to 8748.

Solution

2 3

1 2

T 12 36 T 3 and 3

T 4 T 12

r 3 (the sequence is geometric)

== ==

∴ =

4, 3 and T 8 e but 748 (we know not the posi the a tion o ct r ua l term’s valu value) n a r == = n

1 T n

n ar − = (state the formula)

1 8748 4(3)n− ∴ = (substitute 4, 3 and T 8748 n a r == = )

1

1

7 1

8748 (3) 4

2187 (3)

3 3

7 1

8

n

n

n

n

n

−

−

−

∴ =

∴ =

∴ =

∴ = −

∴ =

∴T 8748 8 =

9

EXAMPLE 8

kk k +1 ; 1 ; 2 5 − − are the first three terms of a geometric sequence.

Calculate the value of k and hence determine the possible sequences.

Solution

2 3

1 2

2

2 2

2

T T

T T

12 5

1 1

( 1) (2 5)( 1)

2 12 3 5

0 6

0 ( 3)( 2)

3 or 2

k k

k k

k kk

kk kk

k k

k k

k k

=

− − ∴ = + −

∴ − = − +

∴ − + = − −

∴ = − −

∴ = − +

∴ = = −

For k = 3: T 1 31 4 1 = += += k For k = −2 : T 1 21 1 1 = += k − + = −

T 1 31 2 2 = k − = − = T 1 21 3 2 = k − = − − = −

T 2 5 2(3) 5 1 3 = k − = − =

The sequence is therefore: The sequence is therefore:

4 ; 2 ;1; ……… −− − 1; 3 ; 9 ; ………

EXERCISE 2

1. Determine the general term for each of the following geometric sequences:

(a) 1 2; 1; ;….. 2 − (b) 2;8;32;….. (c) 2 ; 2; 6;…..

3

− −

(d) 1 ; 0, 2 ; 0,04;…..

2. Determine the 9th term for each of the following geometric sequences:

3. (a) Which term of the geometric sequence 2;6;18;…. is equal to 4374?

(b) Determine the number of terms in the sequence 33 3 ; ; ;….;192

84 2

− −

4. Consider the following sequence: 0,625 ; 1, 25 ; 2,5 ; 5 ; …..

(a) Determine the value of the 10th term.

(b) Determine which term will equal 80.

5. Consider the following sequences:

sequence 1: 2 ; 2 2 ; 3 2 ; ……. sequence 2: 2 ; 2 ; 2 2 ; ………

(a) Show that sequence 1 is arithmetic.

(b) Show that sequence 2 is geometric.

(c) Which term of sequence 1 will be equal to 200 ?

(d) Which term of sequence 2 will be equal to 256 ?

(a) 128; 64; 32;….. (b) 0, 25; 0,5; 1;….. (c) 4 1 ; 1 ; 4 ;…..

9 3

(d) 2 3 ; 1; ;…..

3 2 − (e)

6 2 T 3

3

k

k

− = (f)

1 1 T 2400

2

k

k

− =

T 2 5 2( 2) 5 9 3 = k − = − − = −

10

6. xx x − 4 ; 2 ; 3 1 ; … + + are the first three terms of a geometric sequence.

Determine the sequence if x is positive.

7. t tt +1 ; 1 ; 1 5 ; … − − are the first three terms of a geometric sequence.

(a) Determine the numerical value of t where t ≠ 0.

(b) Determine the sequence.

(c) Determine the 10th term.

(d) Which term equals 2 10

3

?