© 2008  Rasmus ehf    og Jóhann Ísak Pétursson

Sequences and Series

Lesson 2

Geometric Progression GP

A sequence in which each term is a fixed multiple of the previous term is called a geometric sequence or Geometric Progression ( GP ). In other words there is always the same ratio between consecutive terms.

where k is a fixed number. For example in the sequence   100, 50, 25, 12½, 6¼, 3⅛, . . . .   k equals  a half.

Remember the dots show that this is an infinite sequence.

If we  write this sequence term by term we can see the rule for the n th term:

     a1 = 100

a2 = 100∙½ = 50

a3 = 100∙(½)2 = 25

a4 = 100∙(½)3 = 12½

a5 = 100∙(½)4 = 6¼

          

an = 100∙(½)n-1

The following rule holds for a GP:

an = a1∙kn-1

 

Example 1

The first two terms in a GP are 8 and  6. Find the first four terms of the sequence.

First we find the ratio k.

  

Now we can find the third and fourth terms.

a1 = 8

a2 = 8∙¾ = 6

a3 = 8∙(¾)2 = 4½

   a4 = 8∙(¾)3 = 3⅜

Example 2

The third and fourth terms in a GP are  2 and 4. Find the first six terms.

First find the ratio k.

   k = 4/2 = 2

We use this value of k and one of the given terms , for example  a4 , in the rule for a GP to find a1

   a4 = a1∙k3

    4 = a1∙23 = a1∙8

  a1 = 4/8 = ½

Now it's easy to find the rest of the sequence:.

   a2 = ½∙2 = 1

   a3 = ½∙22 = 2

   a4 = ½∙23 = 4

   a5 = ½∙24 = 8

   a6 = ½∙25 = 16

Example 3

In a GP  a3 = 128 and a6 = 2. Find the first six terms.

We need to multiply  a3 = 128 three times by k to get  a6 = 2.

   128∙k3 = 2

          k3 = 2/128 = 1/64

            = ¼

Now we can use the rule to find a1.

      a3 = a1∙k2

   128 = a1∙(¼)2

      a1 = 128/(¼)2 = 128∙16 = 2048

The rest of the sequence follows.

   a2 = 2048∙¼ = 512

   a3 = 2048∙(¼)2 = 128

   a4 = 2048∙(¼)3 = 32

   a5 = 2048∙(¼)4 = 8

   a6 = 2048∙(¼)5 = 2

Example 4

Given a GP with  a3 = 250 and a5 = 62½. Find the ratio k and the first four terms.

We need to multiply 250 twice  (5 – 3 = 2)  by k to get  62½.

  250∙k2 = 62½

         k2 = 62½/250 = ¼

 

          k = ½

In this case there are two possible values for k, a positive and a negative. We find a2 by dividing by k and  a1 by dividing by k2  which is always positive.

  a2 = 250/(½) = 500

  a1 = 500/½ = 1000

This means we get two possibilities for all the even numbered terms in the sequence.

  a1 = 1000

  a2 = 500

  a3 = 250

  a4 = 125

Now let's look at further examples of sequences that obey the rule  an = a1∙kn−1.

Look at the example.

a1 = 12

a2 = 3

a3 = 0.75

a4 = 0.1875

a5 = 0.046875

We see that k = 3/12 = ¼  and the terms decrease rapidly. The sequence can be shown graphically .

As n gets bigger the numbers get rapidly nearer and nearer to zero. The n axis is a horizontal asymptote. Sequences that behave in this way are called convergent sequences. This sequence converges to zero.

A GP is convergent if the ratio k is between 1 and  -1 because multiplication by k makes each term smaller than the previous term and eventually tends to zero.

Rule: A GP is convergent if  −1 < k < 1

A sequence of numbers where each number is a percentage of the previous number is a GP. An increase of   10% means an increase from 100 to110 so the ratio for a sequence with
a  10% increase between terms is   k = 110/100 = 1.1. In the same way, a decrease by  10% means that 100 decreases to 90 so that the ratio   k = 90/100 = 0.9.

 

Here is a list of values of k for different percentages.

      1% increase,  k = 1.01

      5% increase, k = 1.05

    10% increase, k = 1.10

    25% increase, k = 1.25

    90% increase, k = 1.90

  100% increase, k = 2.00

  200% increase, k = 3.00

In general:

p% increase, k = (1 + p/100)

p% decrease, k = (1 − p/100)

 

Example 5

Steve paid  3000 pounds for a car. He borrowed the money at 10 % annual interest without having to repay the loan for five years. The value of the car decreases by 20% each year. How much did it cost Steve to own the car at the end of the five year period if other costs are ignored?
 

First look at the increase because of the interest he has to pay on the loan.

   Original loan, a0 = 3000 pounds.

          After 1 year, a1 = 3000∙1.10 = 3300 pounds.

          After 2 years, a2 = 3300.∙1.10 = 3630 pounds.

          After 3 years, a3 = 3630.∙1.10 = 3993 pounds.

          After 4 years, a4 = 3993.∙1.10 = 4392 pounds.

          After 5 years, a5 = 4392∙1.10 = 4831 pounds.

  Total interest:

       4831 – 3000 = 1831 pounds.

Devaluation of the car:

  Original value a0 = 3000 pounds.

          After 1 year, a1 = 3000∙0.80 = 2400 pounds.

          After 2 years, a2 = 2400∙0.80 = 1.920 pounds.

          After 3 years, a3 = 1920∙0.80 = 1.536 pounds.

          After 4 years, a4 = 1536∙0.80 = 1.228 pounds.

          After 5 years, a5 = 1228∙0.80 = 983  pounds.

  Total loss:

        3000 – 983 = 2017 pounds.

 Total cost         1831+ 2017= 3848 pounds

From the above example we see that when we can calculate compound interest by using the rule   an = a1∙kn−1 
but we need to take care to remember that  a1 represents the value with interest after the first year.

 If, however we use  a0 to represent the original loan or capital then the formula is:

an = a0∙kn

We can also use this formula for  decreases in value, in which case  k < 1.

 

Example 6

Sally deposites 100 pounds in a bank that pays 8% annual interest. How much will she have after 10 ár?

We use the rule a10 = a0∙k10 where  a0 = 100 and k = 1.08

 a10 = 10.000∙1.0810 = 216 pounds.

Example 7

Sally knows of another bank that pays  10% annual interest. If she deposites money in this bank how long will it take to double in value?

   2 = 1∙1.10n

Here we need to use logs to find the index n..

   ln 2 = ln 1.1n

      ln 2 = n∙ln 1.1

Using the rule  ln an = n∙ln a 

       n = ln 2 / ln 1.1 ≈ 7.3 years.

 

Practise these methods then try  Quiz 2  on Sequences and Series  
Remember to use the checklist to keep track of your work.