Arithmetic Progression (A.P)

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 1^st term from the second term or the second term from the third term.

It is important to note 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 4.2.1:

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

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

Solution:

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

T_n = a + (n – 1) d

T₁₂ = -3 + (12 – 1) × 4

= -3 + 11 × 4

= -3 + 44

T₁₂ = 41 

ii. T₂₅ = -3 + (25 – 1) × 4

= -3 + 24 × 4

= -3 + 96

T₂₅ = 93

iii. Let the last term be L

T_n = L = a + (n -1) × d

145 = -3 + (n – 1) × 4

145 = -3 + 4n – 4  

145 = 4n – 7

4n = 152 (divide both sides by 4)

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

n = 38

Example 4.2.2:

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

(i) The common difference 
(ii) The first term of the sequence
(iii) The 61st term of the sequence

Solution

(i) using T_n = a + (n -1)d

Let T₁₁ = a + (11 – 1)d and T₄ = a + (4 – 1)d

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

\( \scriptsize \therefore \) T₁₁ – T₄ = 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 11^th term is \( \scriptsize 2 \frac{24}{25} \! \) times the fourth term

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

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

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

substitute T₄ into the equation

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

substitute T₁₁ into the equation

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

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

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

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

open the bracket

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

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

collect like terms

⇒ \( \frac{74a}{25}\: – \: \frac{a}{1} = \frac{70}{1} \: – \: \frac{1554}{25} \)

find the L.C.M

⇒ \( \frac{74a \: – \: 25a}{25} = \frac{1750 \: – \: 1554}{25} \)

⇒ \( \frac{49a}{25} = \frac{196}{25}\)

multiply both sides by 25

⇒ \( \frac{49a}{25} \: \scriptsize \: \times \: 25 = \normalsize \frac{196}{25} \scriptsize \: \times \: 25\)

49a = 196

divide both sides by 49

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

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

a = 4

(iii) T_n = a + (n -1)d

To find the 61^st term, make n = 61

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

substitute a, d and n into the equation

T₆₁ = a + (61 – 1) × 7

= 4 + 60 × 7

= 4 + 420

T₆₁ = 424

Example 4.2.3:

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

Solution: 

let T₇ = a + 6d and T₉= a + 8d

⇒ \( \frac{T_7}{T_9} = \frac{5}{8} \)

cross multiply

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

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 4.2.4:

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 T₇ = 18

T₇ = 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

T₂ = -12 + (2 – 1)5
= -12 + 5 = -7

3rd term

T₃ = -12 + (3 – 1)5
= -12 + 10 = -2

4th term

T₄ = -12 + (4 – 1)5
= -12 + 15 = 3

5th term

T₅ = -12 + (5 – 1)5
= -12 + 20 = 8

6th term

T₆ = -12 + (6 – 1)5
= -12 + 25 = 13

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

Example 4.2.5:

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 = \( \frac{2\:+\:5\:+\:8}{3} \scriptsize = 5 \)

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

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

= \( \scriptsize 2x \; – \; \normalsize \frac{3}{2} \scriptsize y\)