### What is Arithmetic Progression?

Arithmetic Progression is a sequence in which the difference between consecutive terms is constant. This constant is known as the common difference.

The common difference, d, can be found by subtracting the 1st term from the second term or the second term from the third term.

The general form of an A.P is given as

a, a + d, a + 2d, a + 3d, a + 4d…

where a = 1st term and d = common difference.

It is important to that the coefficient of “d” is always one less than the number of the term. The nth term is denoted by

\(\scriptsize T_n = a \: + \: (n\:-\:1)d \)

Example 1

The first and last terms of an A.P are -3 and 145 respectively. If the common difference is 4, find the

**i.** 12th term**ii. **25th term **iii. **The number of terms in the A.P

**Solution:**

**i.** a = -3, d = 4, n = 12

T_{n} = a + (n – 1) d

T_{12} = -3 + (12 – 1) Ã— 4

= -3 + 11 Ã— 4

= -3 + 44

T_{12} = 41

**ii. **T_{25} = -3 + (25 – 1) Ã— 4

= -3 + 24 Ã— 4

= -3 + 96

T_{25} = 93

**iii.** Let the last term be L

T_{n} = L = a + (n -1) Ã— d

145 = -3 + (n – 1) Ã— 4

145 = -3 + 4n – 4

145 = 4n – 7

4n = 152 (divide both sides by 4)

n = \( \frac {152}{4} \)

n = 38

Example 2:

The difference between the 11th term and the 4th term of an A.P is 49 and if the 11th term is \( \scriptsize 2 \frac{24}{25} \! \) times the fourth term. Determine.Â

**(i)** The common differenceÂ

**(ii)** The first term of the sequence

**(iii)** The 61st term of the sequence

**Solution**

**(i)** using T_{n} = a + (n -1)d

Let T_{11} = a + (11 – 1)d and T_{4} = a + (4 – 1)d

T_{11} = a + (10)d ……..(1)

T_{4} = a + (3)d ……….(2)

The difference between the 11th term and the 4th term of an A.P is 49

\( \scriptsize \therefore \) T_{11} – T_{4 }= 49 …… (3)

**Substitute equations (1) and (2) into equation (3)**

\( \scriptsize \therefore \) a + 10d – (a + 3d) = 49

**open the brackets**

a + 10d – a – 3d = 49

**collect like terms**

~~a~~ – ~~a~~ + 10d – 3d = 49

7d = 49

**divide both sides by 7**

d = \( \frac {49}{7} \)

d = 7

**(ii)** If the 11th term is \( \scriptsize 2 \frac{24}{25} \! \) times the fourth term

â‡’ \( \scriptsize \therefore T_{11} = T_4 \: \times \: 2 \frac{24}{25} \) …… (4)

Substitute equations (1) and (3) into equation (4)

â‡’ \( \scriptsize T_{11} = T_4 \: \times \: 2 \frac{24}{25} \)

**substitute T _{4} into the equation**

â‡’ \( \scriptsize T_{11} = (a \: + \: 3d)\: \times \: \normalsize \frac{74}{25} \)

**substitute T _{11} into the equation**

â‡’ \( \scriptsize a\: + \: 10d = (a \: + \: 3d)\: \times \: \normalsize \frac{74}{25} \)

**substitute the value of d into the equation ( d = 7)**

â‡’ \( \scriptsize a\: + \: (10 \: \times \: 7) = (a \: + \: (3 \: \times \: 7))\: \times \: \normalsize \frac{74}{25} \)

â‡’ \( \scriptsize a\: + \: 70 = (a \: + \: 21)\: \times \: \normalsize \frac{74}{25} \)

**open the bracket**

â‡’ \( \scriptsize a\: + \: 70 = a \: \times \: \normalsize \frac{74}{25} \scriptsize \: + \: 21 \: \times \: \normalsize \frac{74}{25} \)

â‡’ \( \scriptsize a\: + \: 70 = \normalsize \frac{74a}{25} \: + \: \normalsize \frac{1554}{25} \)

**collect like terms**

â‡’ \( \frac{74a}{25}\: – \: \frac{a}{1} = \frac{70}{1} \: – \: \frac{1554}{25} \)

**find the L.C.M**

â‡’ \( \frac{74a \: – \: 25a}{25} = \frac{1750 \: – \: 1554}{25} \)

â‡’ \( \frac{49a}{25} = \frac{196}{25}\)

**multiply both sides by 25**

â‡’ \( \frac{49a}{25} \: \scriptsize \: \times \: 25 = \normalsize \frac{196}{25} \scriptsize \: \times \: 25\)

49a = 196

**divide both sides by 49**

â‡’ \( \frac{49a}{49} = \frac{196}{49} \)

a = \( \frac{196}{49}\)

a = 4

**(iii)** T_{n} = a + (n -1)d

To find the 61st term, make n = 61

from the question i and ii….. a = 4, d = 7

substitute a, d and n into the equation

T_{61} = a + (61 – 1) Ã— 7

= 4 + 60 Ã— 7

= 4 + 420

T_{61} = 424

Example 3:

The first term of an A.P is -8. The ratio of the 7th term to the 9th is 5:8. Calculate the common difference of the Progression. (WAEC)

**Solution: **

let T_{7} = a + 6d and T_{9}= a + 8d

â‡’ \( \frac{T_7}{T_9} = \frac{5}{8} \)

**cross multiply**

i.e \( \frac{a \:+\: 6d}{a \:+\: 8d} \scriptsize = \normalsize \frac{5}{8}\)

8(a + 6d) = 5(a + 8d)

8a + 48d = 5a + 40d

8a – 5a = 40d – 48d

3a = -8d

From the question the first term a = -8

\( \scriptsize \therefore 3 \: \times \: -8 = \: -8d \) \( \scriptsize -24 = -8d \)d = \( \frac{-24}{-8} \)

d = 3

Example 4:

Insert 5 arithmetic means between 18 and -12

**Solution**

Total number of terms = 7 ( the first term is -12, last term 18 and 5 terms in between, making 7)

a = -12 and T_{7} = 18

T_{7} = a + (7 – 1)d

18 = a + 6d

18 = -12 + 6d

6d = 30

d = 5

the 5 arithmetic means are (1st term = -12)

using T_{n} = a + (n – 1)d

**2nd term**

T_{2} = -12 + (2 – 1)5

= -12 + 5 = -7

**3rd term**

T_{3} = -12 + (3 – 1)5

= -12 + 10 = -2

**4th term**

T_{4} = -12 + (4 – 1)5

= -12 + 15 = 3

**5th term**

T_{5} = -12 + (5 – 1)5

= -12 + 20 = 8

**6th term**

T_{6} = -12 + (6 – 1)5

= -12 + 25 = 13

i.e -7, -2, 3, 8, 13

Example 5:

Find the arithmetic mean of each of the following

**(i)** -14 and 16

**(ii)** 2, 5, 8

**(iii)** x + y and 3x – 4y

**Solution**

**(i)** A.M = \( \frac{-14 + 16}{2} \scriptsize = \normalsize \frac{2}{2} \scriptsize = 1 \)

**(ii) **2, 5, 8 the A.M = the middle value = 5

**(iii)** \( \frac{x + y + 3x – 4y}{2} \scriptsize = \normalsize \frac{4x – 3y}{2} \)

i.e A.M = \( \frac {1}{2} \scriptsize \left ( 4x – 3y \right) \)

â‡’ \( \scriptsize 2x \; – \; \normalsize \frac{3}{2} \scriptsize y\)

Well explain