**Contents**

## Arithmetic progressions

Number sequences that have a **constant difference** between successive terms are called arithmetic progressions (AP).

For example:

- 1, 3, 5, 7, 9, … is an arithmetic progression with a common difference of 2.
- a, a+d, a+2d, a+3d, … is an arithmetic progression with a common difference of d.
- 10, 20, 30, 40, … is an arithmetic progression with a common difference of 10.

In general the first term of an AP is represented by ‘**a**‘ and the common difference by ‘**d**‘ and the formulae for the n^{th} term of the AP is:

*n ^{th} term = a+(n − 1)d*

### Question 1

Find the 23^{rd} term of the series 1, 5, 9, 13, …..

### Solution 1

*First we calculate the common difference, “d” by subtracting any two adjacent terms in the series.*

d = 5-1

= 4

*The first term, “a” is 1.*

n^{th} term = a+(n−1)d

23^{rd} term = 1+(23-1)(4)

= 1+(22)(4)

= 1+88

= 89

## Last term of an AP

Given the number of terms of a series, we can find the last term of the series using the formula *n ^{th} term = a+(n−1)d* and using using the number of terms in the series as

*n*. Representing the last term by

*l*, this becomes:

*l = a+(n-1)d*

for *n* terms.

### Question 2

Find the last term of the series 1, 5, 9, 13,… given that the series contains 100 terms.

### Solution 2

*First we calculate the common difference, “d” by subtracting any two adjacent terms in the series.*

d = 5-1

= 4

*The first term, “a” is 1.*

last term = a+(n−1)d

last term = 1+(100-1)(4)

= 1+(99)(4)

= 1+396

= 397

## Sum of the terms of an AP

To find the sum S of an AP we multiply the average of all the terms by the number of terms.

Which means that

Substituting *l = a+(n−1)d*

We get

Therefore sum of an AP of *n* terms:

### Question 3

Find the sum of the series 1, 5, 9, 13,… given that the series contains 100 terms.

### Solution 3

## Worked Examples

### Question 4

Determine (a) the 3^{rd}, and (b) the 23^{rd} term of the series 3, 8, 13, 18,…

### Solution 4

a) d = 8-3

= 5

a = 3

n^{th} term = a+(n−1)d

3^{rd} term = 3+(3-1)(5)

= 1+(4)(5)

= 1+20

= 21

b) d = 8-3

= 5

a = 3

n^{th} term = a+(n−1)d

23^{rd} term = 3+(23-1)(5)

= 1+(22)(5)

= 1+110

= 11

### Question 5

The 3^{rd} term of an AP is 23 and the 23^{rd} term is 87. Determine the 20^{th} term.

### Solution 5

n^{th} term = a+(n−1)d

Using the 3^{rd} term: 23 = a+(3−1)d

Using the 23^{rd} term: 87 = a+(23-1)d

23 = a+2d ——(equation 1)

87 = a+22d —–(equation 2)

(eqn 2) minus (eqn 1)

64 = 20d

d = 3.2

Using equation 1: 23 = a+2d

23 = a+2(3.2)

23 = a+6.4

23-6.4 = a

a = 16.6

Hence 19^{th} term = 16.6+(19−1)(3.2)

= 16.6+(18)(3.2)

= 16.6+57.6

= 74.2

### Question 6

Determine the position of the term whose value is 22 in the series 2.5, 4, 5.5, 7…

### Solution 6

a = 2.5

d = 1.5

and the n^{th} term is 22 which means:

22 = a+(n−1)d

22 = 2.5+(n-1)(1.5)

22 = 2.5+1.5n-1.5

22 = 1+1.5n

22-1 = 1.5n

21 = 1.5n

n = 14

Therefore 22 is the 14^{th} term.

### Question 7

Find the sum of the first 42 terms of the series 1, 11, 21, 31,…

### Solution 7

a = 1

d = 10

n = 42

### Question 8

Find the 11th term of the series 8, 14, 20, 26,… .

### Solution 8

a = 8

d = 14-8 = 6

n^{th} term = a+(n-1)d

11^{th} term = 8+(11-1)(6)

= 8+10(6)

= 8+60

= 68

### Question 9

Find the 17th term of the series 11, 10.7, 10.4, 10.1,… .

### Solution 9

a = 11

d = 10.7-11 = -0.3

n^{th} term = a+(n-1)d

17^{th} term = 11+(17-1)(-0.3)

= 11+(16)(-0.3)

= 11-4.8

= 6.2

### Question 10

The seventh term of a series is 29 and the eleventh term is 54. Determine the sixteenth term.

### Solution 10

n^{th} term = a+(n-1)d

7^{th} term => 29 = a+(7-1)d

11^{th} term => 54 = a+(11-1)d

29 = a+6d and 54 = a+10d

Then solve the simultaneous equations to get:

d = 6.25 and a = -8.5

Therefore 16^{th} term = -8.5+(16-1)(6.25)

= -8.5+(15)(6.25)

= 85.25

### Question 11

Find the 15th term of an arithmetic progression of which the first term is 2.5 and the tenth term is 16.

### Solution 11

a = 2.5

10^{th} term = 16

n^{th} term = a+(n-1)d

10^{th} term => 16 = 2.5+(10-1)d

16 = 2.5+9d

13.5 = 9d

d = 1.5

Therefore 15^{th} term = 2.5+(15-1)(1.5)

= 2.5+14(1.5)

= 2.5+21

= 23.5

### Question 12

Determine the number of the term which is 29 in the series 7, 9.2, 11.4, 13.6,… .

### Solution 12

a = 7

d = 2.2

n^{th} term = 29

n^{th} term = a+(n-1)d

29 = 7+(n-1)(2.2)

29 = 7+2.2n-2.2

29 = 4.8+2.2n

29-4.8 = 2.2n

24.2 = 2.2n

n = 11

### Question 13

Find the sum of the first 11 terms of the series 4, 7, 10, 13,… .

### Solution 13

a = 4

d = 3

n = 11

### Question 14

Determine the sum of the series 6.5, 8.0, 9.5, 11.0, … , 32.

### Solution 14

a = 6.5

d = 1.5

l = 32

l = a+(n-1)d

32 = 6.5+(n-1)(1.5)

32 = 6.5+1.5n-1.5

32 = 5+1.5n

32-5 = 1.5n

1.5n = 27

n = 18