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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 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)

  1. P: a, ar,     ar2,        ar3 ………………

 

Denotations in G. P

a    = 1st term

r    = common ratio

Un = nth term

Sn  = sum

 

The nth term of a G. P

The nth term = Un

Un = arn-1

1st term   = a

2nd term = a x r =ar

3rd term = a x r x r = ar2

4th term = a x r x r x r =  ar3

8thterm  =  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 ……….. arn-1

 

Example

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

Solution

a = 5         r  = 10/5  =  2

U9         =   arn-1

U9         =   5 (2) 9-1

=   5 (2)8

=   5 x 256   = 1,280

(b)        Un = arn-1

= 5(2) n-1

Example 2

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

U8  =  -7/32           Un        =  arn-1

-7/32   =  28 (r)8-1

-7/32   =  28r7

-7/32 x 1/28 =

-7

 

896

7/32 x 1/28 = r7

    – 7

32 x 28

7                                       7

r =                              =

 

r = – 0.5

 

Evaluation

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

 

THE SUM OF A GEOMETRIC SERIES

a + ar + ar2 + ar3 + ………………. arn-1

represent a general geometric series where the terms are added.

S =  a + ar + ar2 ………… arn-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.

  1. ½ + ¼ + 1/8 + …………………… as far as 6th term
  2. 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

S6½ (1 – 1/64)

½

S6   = 1 – 1   = 64 – 1   = 63

64       64          64

  1. a = 1, r = 3, n = ? Un = 729

Un = arn-1

729 = 1 x 3n-1             (3n-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 7th 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.

  1. 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 3rd and 6th 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. 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  7th  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 2nd and 4th 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 4th 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

  1. If the 2nd and 5th 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
  2. 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 3rd term of a GP is 360 and the 6th 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

ARITHMETIC PROGRESSION (A. P)

PERCENTAGE ERROR

LOGARITHM OF NUMBERS

MENSURATION

Symmetric Properties of Roots

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