# Arithmetic Progression

An arithmetic progression is a sequence of numbers where the difference between the numbers in the same/Constant.

« Previous Next »## Arithematic progression key Concepts

Given that the following sequence of numbers are the time intervals taken by a cyclist to move from one station to another

1, 4 , 7, 10

Challenges

a) Find the difference in time

b) Find the time a cyclist will taken to reach station 6

c)Therefore find the sum of A.P on the 6 station

Solutions

#### How to find the common difference

The time difference from the sequence is:

a) Difference = Term2 - Term1

To find the difference subtract a term or number which is next in the sequence by a previous one

Difference = 4 - 1

Difference = 3

## Points to note

The sequence below is the pattern that is used to obtain a term

a , a + d, a + 2d , a + 3d

a = is the first term

d = is the common difference

To find the fourth 4th term

a + 3d

1 + 3(3)

1 + 9

Answer: 10

The 4th term is: 10 .Therefore deriving a formula to find nth term

Formula

a + (n - 1)d

a = the first term

n = unknown term

d = the difference

proving if the nth term formula will give us the same answer for the 4th term

a + (n - 1)d

1 + (4 - 1)3

1 + (3)3

Answer: 10

b) The time to reach station 6

a + (n - 1)d

1 + (6 - 1)3

1 + (5)3

1 + 15

Answer: 16

c) Sum of A.P at station 6

#### How to find the Sum of an Arithematic progression

Formula

$Sum\; =\frac{\mathrm{n}}{2}(\mathrm{a}+\mathrm{L})$$Sum\; =\frac{\mathrm{n}}{2}(\mathrm{2a}+\mathrm{(n\; -\; 1)\; d})$

n = 6

a = 1

d = 3

Solution:

$Sum\; =\frac{6}{2}(\mathrm{2(1)}+\mathrm{(6\; -\; 1)\; 3})$$Sum\; =3(2+\mathrm{(5)\; 3})$

$Sum\; =3(2+15)$

$Sum\; =3\left(17\right)$

$Sum\; =51$

## Example 1

(a) For the sequence 2, 4, 6, 8, 10, 12, 14, 16 find the 15th term.

(b) For the sequence 14, 17, 20, 23, ... Find the,

(i) 18th term

(ii) Sum of the first 30 terms.

(c) Find the sum 24 terms of 1, 5, 9, 13, 17 ...

(d) The fifth term of an AP is 10 and the tenth term is 20, find the values of a and d, hence find the 20th term.

Solutions

(a)

The formula for finding a given term is nth = a + (n - 1)d we want to find term number 15 so we have n = 15

Find d which is the common difference

d = 4 - 2

d = 2

a = 2, n = 15 , d = 2

15th = a + (n - 1)d

15th = 2 + (15 - 1)2

15th = 2 + (14)2

15th = 2 + 28

Answer: 15th = 30

(b)

(i)

18th = a + (n - 1)d

Sequence = 14, 17, 20, 23

d = 17 - 14

d = 3

a = 14, n = 18, d = 3

18th = 14 + (18 - 1)3

18th = 14 + (17)3

18th = 14 + 51

Answer: 18th = 65

(ii)

Sum of first 30 terms

Formula

$Sum\; =\frac{\mathrm{n}}{2}(\mathrm{2a}+\mathrm{(n\; -\; 1)\; d})$

a = 14, n = 30, d = 3

$Sum\; =\frac{30}{2}(\mathrm{2(14)}+\mathrm{(30\; -\; 1)\; 3})$

$Sum\; =15(28+\mathrm{(29)\; 3})$

$Sum\; =15(28+87)$

$Sum\; =15\left(115\right)$

$Sum\; =1725$

Answer $Sum30\; =1725$

(c)

Sequence is 1, 5, 9, 13, 17

Sum of the first 24 terms

d = 5 - 1

d = 4

a = 1, n = 24, d = 4

$Sum\; =\frac{\mathrm{n}}{2}(\mathrm{2a}+\mathrm{(n\; -\; 1)\; d})$

$Sum\; =\frac{24}{2}(\mathrm{2(1)}+\mathrm{(24\; -\; 1)\; 4})$

$Sum\; =12(2+\mathrm{(23)\; 4})$

$Sum\; =12(2+92)$

$Sum\; =12\left(94\right)$

$Sum\; =1128$

Answer: $Sum24\; =1128$

(d)

The formula for term of an AP is nth = a + (n - 1)d

Fifth term is 10

Meaning term number 5 is 10, so n = 5, then term number 5 is 10. So we have;

T5 = a + (5 - 1)d

T5 = a + (4)d

a + 4d = 10 ....(i)

The tenth term is 20

Meaning term number 10 is 20, so n = 10, then term number 10 is 20. So we have;

T10 = a + (10 - 1)d

T10 = a + (9)d

a + 9d = 20 ....(ii)

Solve equation i and ii simultaneously

a + 4d = 10 ....(i)

a + 9d = 20 ....(ii)

a = 10 - 4d ....(i)

substitute the value a into the equaction (ii)

a + 9d = 20 ....(ii)

10 - 4d + 9d = 20 ....(ii)

- 4d + 9d = 20 - 10

5d = 10

$\frac{\mathrm{5d}}{5}=\frac{\mathrm{10}}{5}$

Answer: $d=2$

substitute d into the equation below

a = 10 - 4d ....(i)

a = 10 - 4(2)

a = 10 - 8

Answer: a = 2

Find the 20th term

nth = a + (n - 1)d

a = 2 , n = 20 , d = 2

nth = a + (n - 1)d

20th = 2 + (20 - 1)2

20th = 2 + (19)2

20th = 2 + 38

Answer: 20th = 40

## Example 2

For the sequence 2, 5, 8, 11, 14 ... find

(i) An expression for the nth term

(ii) The 9th term using an expression in (i)

(iii) The 100th term using an expression in (i)

(iv) An expression for the sum of the first nth terms

(v) The sum of the first 40 terms using an expression in (i)

Solutions

(i)

Formula nth = a + (n - 1)d

d = 5 - 2

d = 3

a = 2, n = ? d = 3

nth = 2 + (n - 1)3

nth = 2 + 3n - 3

nth = 3n + 2 - 3

Answer: nth = 3n - 1

(ii)

9th term

nth = 3n - 1

9th = 3(9) - 1

9th = 27 - 1

Answer: 9th = 26

(iii)

100th term

nth = 3n - 1

100th = 3(100) - 1

100th = 300 - 1

Answer: 100th = 299

(iv)

Formula $Sum\; =\frac{\mathrm{n}}{2}(\mathrm{2a}+\mathrm{(n\; -\; 1)\; d})$

a = 2 , d = 3, n = ?

$Sum\; =\frac{\mathrm{n}}{2}(\mathrm{2(2)}+\mathrm{(n\; -\; 1)\; 3})$

$Sum\; =\frac{\mathrm{n}}{2}(4+\mathrm{3n\; -\; 3})$

$Sum\; =\frac{\mathrm{n}}{2}\left(\mathrm{3n\; +\; 4\; -\; 3}\right)$

$Sum\; =\frac{\mathrm{n}}{2}\left(\mathrm{3n\; +\; 1}\right)$

Answer: $Sth\; =\frac{{\mathrm{3n}}^{2}}{2}+\frac{\mathrm{n}}{2}$

(iv)

$Sth\; =\frac{{\mathrm{3n}}^{2}}{2}+\frac{\mathrm{n}}{2}$

n = 40th

$S40\; =\frac{{\mathrm{3(40)}}^{2}}{2}+\frac{40}{2}$

$S40\; =\frac{\mathrm{3(1600)}}{2}+20$

$S40\; =\frac{\mathrm{4800}}{2}+20$

$S40\; =\mathrm{2400}+20$

Answer: $S40\; =2420$

## Example 3

Given that the 11th term of an arithmetic progression is 43 and that the first term is 3,find the

a) common difference

b) 4 th term

Solution

First term a = 3

11th term = 43

Formula: Tn = a + (n - 1) d

a)

43 = 3 + (11 - 1) d

43 = 3 + (10) d

43 - 3 = (10) d

40 = (10) d

$\frac{\mathrm{40}}{10}=\frac{\mathrm{10(d)}}{10}$

Common difference: d = 4

b)

Formula: Tn = a + (n - 1) d

a = 3, n = 4, d = 4

T4 = 3 + (4 - 1) 4

T4 = 3 + (3) 4

T4 = 3 + 12

Fourth term T4 = 15