Geometric Progression SS1 Mathematics Lesson Note

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

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 + ar³ ………… arⁿ-¹ eqn 1

Multiply through r

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

subtract eqn 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  

S6=  ½ (1 – 1/64)

½ 

S6   = 1 – 1   = 64 – 1   = 63

    64     64          64

2. 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

SUM OF G. P. TO INFINITY

The 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 terms 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.

100. n = 100.

S = a (1 – rⁿ)

          1 – r

S =  1 (1-(1/2)100)  

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

In general,

S = a (1-rn). =  a(1-0)  =   a__

        1-r       1-r         1 – r

S = a

     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- ⅔

S∞  =  45 ÷ 1/3

          45 x 3/1

         =  135

ASSIGNMENT 

1. The 3rd term of a GP is 360 and the 6th term is 1215. Find the
2. Common ratio        
3. First term        

iii. Sum of the first four terms

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