GEOMETRIC PROGRESSION

Subject: 

MATHEMATICS

Term:

FIRST TERM

Week:

WEEK 4

Class:

SS 2

Topic:

GEOMETRIC PROGRESSION

 

Previous lesson: 

The pupils have previous knowledge of

ARITHMETIC PROGRESSION (A. P)

that was taught as a topic in the previous lesson

 

Behavioural objectives:

At the end of the lesson, the learners will be able to

 

• Say the definition of Geometric Progression
• Write out the signs that show the denotations of Geometric progression
• Calculate the nth term of a G. P.
• Calculate and write out the sum of Geometric series
• Solve simple sums on the sum of G. P. to infinity
• Geometric mean

 

Instructional Materials:

• Wall charts
• Pictures
• Related Online Video
• Flash Cards

 

 

Methods of Teaching:

• Class Discussion
• Group Discussion
• Asking Questions
• Explanation
• Role Modelling
• Role Delegation

 

Reference Materials:

• Scheme of Work
• Online Information
• Textbooks
• Workbooks

 

Content:

 

WEEK FOUR

TOPIC: GEOMETRIC PROGRESSION

CONTENT

• Definition of Geometric Progression
• Denotations of Geometric progression
• The nth term of a G. P.
• The sum of Geometric series
• Sum of G. P. to infinity
• Geometric mean

Definition of G. P

The sequence 5, 10, 20, 40 has a first term of 5 and the common ratio

Between the terms are 2 e.g. (¹⁰/₅ or ⁴⁰/_2o = 2).

A sequence in which the terms either increase or decrease in a common ratio is called a Geometric Progression

(G. P)

G. P: a, ar, ar², ar³ ………………

Denotations in G. P

a = 1^st term

r = common ratio

U_n = nth term

S_n = sum

The nth term of a G. P

The nth term = Un

Un = ar^n-1

1^st term = a

2^nd term = a x r =ar

3^rd term = a x r x r = ar²

4^th term = a x r x r x r = ar³

8^thterm = a x r x r x r x r x r x r x r = ar⁷

nth term = a x r x r x r x ……….. ar^n-1

Example

Given the GP 5, 10, 20, 40. Find its (a) 9^th term (b) nth term

Solution

a = 5 r = 10/5 = 2

U₉ = ar^n-1

U₉ = 5 (2) ^9-1

= 5 (2)⁸

= 5 x 256 = 1,280

(b) U_n = ar^n-1

= 5(2) ^n-1

Example 2

The 8^th term of a G.P is -7/32. Find its common ratio if it first term is 28.

U₈ = -7/32 U_n = ar^n-1

-7/32 = 28 (r)^8-1

-7/32 = 28r⁷

-7/32 x 1/28 =

–⁷/₃₂ x ¹/₂₈ = r⁷

-7

896

– 7

32 x 28

7 7

r = =

r = – 0.5

Evaluation

1. The 6^th term of a G.P is 2000. Find its first term if its common ratio is 10.

2. Find the 7^th term and the nth term of the progression 27, 9 , 3, …

THE SUM OF A GEOMETRIC SERIES

a + ar + ar² + ar³ + ………………. ar^n-1

represent a general geometric series where the terms are added.

S = a + ar + ar² ………… ar^n-1 eqn 1

Multiply through r

rs = ar + ar² + ar³ ………. arⁿ ……… eqn 2

subtracteqn 2 from 1

S – rs = a – arⁿ

S (1 – r) =a(1-rⁿ)

1 – r 1-r

S.=a ( 1 – rⁿ) r < 1

1 – r

Multiply through by -1 or subs. eqn. 1 from e.g. 2

rs – s = arⁿ – a

S (r – 1) =a(rⁿ – 1)

r – 1 r – 1

S = a(rⁿ-1)

r -1 for r > 1

Example:

Find the sum of the series.

a. ½ + ¼ + 1/8 + …………………… as far as 6^th term

b. 1 + 3 + 9 + 27 + …………………. 729

Solution

a = ½

r = ½ (r = ¼ ÷ ½ = ½)

∴r< 1

S = a (1-rⁿ)

1 – r

S₆ = [½ (1 – (½)⁶]

1 – ½

S₆= ½ (1 – ¹/₆₄)

½

S₆ = 1 – 1 = 64 – 1 = 63

64 64 64

2. a = 1, r = 3, n = ? U_n = 729

U_n = ar^n-1

729 = 1 x 3^n-1 (3^n-1 = 3ⁿ x 3^-1)

729 = 3n

3

3ⁿ = 3 x 729

3ⁿ = 31 x 36

3ⁿ = 3⁷

∴ n = 7

S = a(rⁿ-1)

r – 1

S = a(3⁷ – 1) = 2187 – 1

3 – 1 2

2186 = 1093

2

Evaluation: Find the sum of the series 40, -4, 0.4 as far as the 7^th term.

SUM OF G. P. TO INFINITY

Sum of G. P to infinity is only possible where r is < 1.

Where r is > 1 there is no sum to infinity.

Example:

1. Find the sum of G. P. 1 + ½ + ¼ + …………………… (a) to 10 terms (b) to 100 terms. Hence deduce the sum of the series (formula) if it has a very large no. of term or infinity.

(a) a = 1 r = ½

n = 10

S = a (1-rⁿ)

1-r

S = 1(1-(¹/₂)¹⁰) = 1(1-0.0001)

1- ½ 1/2

 

2 (1 – 0.001)

2 – 0.002 = 1.998.

b. n = 100.

S = a (1 – rⁿ)

1 – r

S = 1 (1-(¹/₂)¹⁰⁰) = 1(1- (¹/₂)¹⁰)¹⁰

1 – ½ ½

1 (1-(0.001)¹⁰

½

1 (1)

½ = 2

Therefore (1/2)¹⁰⁰ tend to 0 (infinity).

In general,

S = a (1-rⁿ)= a(1-0) = a__

1-r 1-r 1 – r

∴ S_∞= a__ = n → ∞

1 – r

Example 2:

Find the sum of the series 45 + 30 + 20 + ……………… to infinity.

a = 45, r = ²/₃, n = infinity

S∞ = a S = 45__

1 – r 1- ²/₃

S∞ = 45 ÷ ¹/₃

45 x ³/₁

= 135

Evaluation

1. The sum to infinity of a Geometric Series is 100. Find the first term if the common ratio is –¹/₂.

2. The 3^rd and 6^th term of a G. P. are 48 and 14²/₉ respectively, write down the first four terms of the G. P.

3. The sum of a G. P. is 100 find its first term if the common ratio is 0.8.

GEOMETRIC MEAN

If three numbers such as x , y and z are consecutive terms of a G.P then their common ratio will be

y =z

x y

y² = xz

y = xz

The middle value , y is the geometric mean (GM). We can conclude by saying that the GM of two numbers is the positive square root of their products.

Example

Calculate the geometric mean of I. 3 and 27 II. 49 and 25

4

Solution

1. G.M of 3 and 27 II. G.M of 49 and 25

= √ 3 x 27 4

= √ 81 = 49 x 25

= 9 4

= 7 x 5

2

= 35 = 17 1/2

2

Example

The first three terms of a GP are k + 1, 2k – 1, 3k + 1. Find the possible values of the common ratio.

Solution

The terms are k + 1, 2k – 1, 3k + 1

2k -1 =3k + 1

k + 1 2k – 1

(2k-1)(2k-1) = (k+1)(3k+1)

4k²-2k-2k +1 = 3k² +k+3k + 1

4k²– 4k +1 = 3k² +4k + 1

4k² – 3k² – 4k – 4k + 1-1 = 0

k² -8k = 0

k(k-8) = 0

k = 0 or k – 8 = 0

k = 0 or 8

The common ratio will have two values due to the two values of k

When k=0 when k= 8

K+1 = 0+1 =1 k+1 = 8+1 = 9

2k- 1= 2×0 – 1 = -1 2k- 1 = 2×8 – 1 = 15

3k+ 1= 3×0+ 1 = 1 3k+1 = 3×8 +1 = 25

terms are 1 , -1 , 1 terms are 9,15,25

common ratio, r = -1/1 common ratio,r = 15/9

r = -1

EVALUATION

The third term of a G.P. is 1/81. Determine the first term if the common ratio is 1/3.

GENERAL EVALUATION /REVISION QUESTION

1. p – 6, 2p and 8p + 20 are three consecutive terms of a GP. Determine the value of (a) p (b) the common ratio

2. If 1 , x , 1 , y , ….are in GP , find the product of x and y

16 4

3.The third term of a G.P is 45 and the fifth term 405.Find the G.P. if the common ratio r is positive.

4.Find the 7^th term and the nth term of the progression 27,9,3,…

5.In a G.P, the second and fourth terms are 0.04 and 1 respectively. Find the (a) common ratio (b) first term

WEEKEND ASSIGNMENT

1. In the 2^nd and 4^th term of a G.P are 8 and 32 respectively, what is the sum of the first four terms. (a) 28 (b) 40 (c) 48 (d) 60

2. The sum of the first five term of the G.P. 2, 6, 18, is (a) 484 (b) 243 (c) 242 (d) 130

3. The 4^th term of a GP is -2/3 and its first term is 18 what is its common ratio. (a) ½ (b) ¹/₃

(c) ^-1/₃ (d) ^-1/₂

4. If the 2^nd and 5^th term of a G. P. are -6 and 48 respectively, find the sum of the first four terms: (a) -45 (b) -15 (c) 15 (d) 33

5. Find the first term of the G.P. if its common ratio and sum to infinity – ³/₃ and respectively (a) 48 (b) 18 (c) 40 (d) -42

THEORY

1.The 3^rd term of a GP is 360 and the 6^th term is 1215. Find the

(i) Common ratio (ii) First term (iii) Sum of the first four terms

1b. If (3- x) + (6) + (7- 5x) is a geometric series, find two possible values for

(i) x (ii) the common ratio, r (iii) the sum of the G.P

2.The first term of a G. P. is 48. Find the common ratio between its terms if its sum to infinity is 36.

Reading Assignment

New General Mathematics SSS2

 

 

Conclusion

The class teacher wraps up or concludes the lesson by giving out short notes to summarize the topic that he or she has just taught.

The class teacher also goes round to make sure that the notes are well copied or well written by the pupils.

He or she makes the necessary corrections when and where the needs arise.