Lesson 1, Topic 2
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# Arithmetic Progression (A.P)

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### 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

Tn = a + (n – 1) d

T12 = -3 + (12 – 1) Ã— 4

= -3 + 11 Ã— 4

= -3 + 44

T12 = 41

ii. T25 = -3 + (25 – 1) Ã— 4

= -3 + 24 Ã— 4

= -3 + 96

T25 = 93

iii. Let the last term be L

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

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

T11 = a + (10)d ……..(1)

T4 = a + (3)d ……….(2)

The difference between the 11th 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

aa + 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 T4 into the equation

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

substitute T11 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) Tn = 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

T61 = a + (61 – 1) Ã— 7

= 4 + 60 Ã— 7

= 4 + 420

T61 = 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 T7 = a + 6d and T9= 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 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 Tn = 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 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‡’ $$\scriptsize 2x \; – \; \normalsize \frac{3}{2} \scriptsize y$$

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