**GEOMETRIC PROGRESSION**

**CONTENT**

Table of Contents

- 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__

__Definition of G. P__

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

Between the term is 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)

- P: a, ar, ar
^{2}, ar^{3}………………

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 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 = ar^{2}

4^{th} term = a x r x r x r = ar^{3}

8^{th}term = a x r x r x r x r x r x r x r = ar^{7}

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

U_{9} = ar^{n-1}

U_{9} = 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.

U_{8} = -7/32 U_{n} = ar^{n-1}

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

-7/32 = 28r^{7}

-7/32 x 1/28 =

-7
896 |

–^{7}/_{32} x ^{1}/_{28} = r^{7}

– 7
32 x 28 |

7 7

r = =

r = – 0.5

__Evaluation__

- The 6
^{th}term of a G.P is 2000. Find its first term if its common ratio is 10. - Find the 7
^{th}term and the nth term of the progression 27, 9 , 3, …

** **

**THE SUM OF A GEOMETRIC SERIES**

a + ar + ar^{2} + ar^{3} + ………………. ar^{n-1}

represent a general geometric series where the terms are added.

S = a + ar + ar^{2} ………… ar^{n-1 }eqn 1

Multiply through r

rs = ar + ar^{2} + ar^{3} ………. ar^{n} ……… eqn 2

subtracteqn 2 from 1

S – rs = a – ar^{n}

S __(1 – r)__ =__a(1-r ^{n})__

1 – r 1-r

**S.= a ( 1 – r^{n})** r < 1

**1 – r**

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

rs – s = ar^{n} – a

S __(r – 1)__ =__a(r ^{n} – 1)__

r – 1 r – 1

S = __a(r ^{n}-1)__

**r -1** for r > 1

__Example:__

Find the sum of the series.

- ½ + ¼ + 1/8 + …………………… as far as 6
^{th}term - 1 + 3 + 9 + 27 + …………………. 729

__Solution__

a = ½

r = ½ (r = ¼ ¸ ½ = ½)

\r< 1

S = __a (1-r ^{n})__

1 – r

S_{6} = [__½ (1 – (½) ^{6}]__

1 – ½

S_{6}= __½ (1 – ^{1}/_{64})__

½

S_{6 } = 1 –__ 1 __ = __64 – 1__ = __63__

64 64 64

- a = 1, r = 3, n = ? U
_{n}= 729

U_{n} = ar^{n-1}

729 = 1 x 3^{n-1} (3^{n-1} = 3^{n} x 3^{-1})

729 = __3n__

3

3^{n} = 3 x 729

3^{n} = 31 x 36

3^{n} = 3^{7}

\ n = 7

S = __a(r ^{n}-1)__

r – 1

S = __a(3 ^{7} – 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:

- 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-r ^{n})__

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.

- n = 100.

S = __a (1 – r ^{n})__

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-r ^{n})__=

__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__

- The sum to infinity of a Geometric Series is 100. Find the first term if the common ratio is –
^{1}/_{2}. - The 3
^{rd}and 6^{th}term of a G. P. are 48 and 14^{2}/_{9}respectively, write down the first four terms of the G. P. - 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

y^{2} = 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**

- 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)

4k^{2}-2k-2k +1 = 3k^{2} +k+3k + 1

4k^{2}– 4k +1 = 3k^{2} +4k + 1

4k^{2 }– 3k^{2 }– 4k – 4k + 1-1 = 0

k^{2} -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**

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

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

- 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 - 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.

See also