geometric sequences & series this week the focus is on using geometric sequences and series to...

Geometric Sequences & Series

This week the focus is on using

geometric sequences and series to

solve problems involving growth and

decay.


CONTENTS

Growth and Decay

Interest Rates

Interest Rates Illustration

Example 1

Example 2

Example 3

Assignment


Growth and Decay

Geometric sequences can be used to solve

problems involving growth or decay at a

constant rate.

Most commonly geometric sequences and

series are used in problems involving

interest rates and population

growth/decay.


Growth and Decay

Interest Rates:

If property values are increasing at a rate of 4%

per annum, the sequence of values increases

as a geometric sequences with a common ratio

of 1.04.

Similarly, if a property decreased by 4% per

annum the common ratio for the geometric

sequence would be 0.96.

The same principal also applies to interest rates

for financial investments and population

changes.


Growth and Decay

Illustration:

In 2000 a property was valued at £100,000. Over

the next 3 years the property increased in

value at a rate of 4% per annum. This gives

the following values of the property:

2000 £100,000

2001 £104,000

2002 £108,160

2003 £112,486.40

a = 100,000 r = 1.04


Example 1:I invest £A in the bank at a rate of interest of

3.5% per annum. How long will it be before I double my investment?

Solution:We can take the growth of the investment as a

geometric sequence, the first term is A and the common ratio is 1.035 because the interest rate is 3.5%.

The question is really asking what term will give us twice the amount we started with – i.e 2A.

This means we are trying to find the value of n which gives 2A. continued on next slide

Geometric Sequences & SeriesSolution continued:Substituting values into the general formula for the nth term of a geometric sequence gives:

A(1.035)n = 2A(1.035)n = 2 by dividing across by A

Take the log of both sides to solve the equation involving powers to get:

log(1.035)n = log 2n log(1.035) = log 2n = log 2/log 1.035 = 20.15

Therefore my money will have doubled after 20.15 years.


Example 2:A car depreciates in value by 15% a year. If it is

worth £11,054.25 after 3 years, what was its new price and when will it be first worth less than £5000.

Solution: We can take the depreciation of the car as a

geometric sequence with first term a (the new price) and the common ratio is 0.85 (100% - 15%).

We know that after 3 years the value is £11,054.25. Although this value is after 3 years, this is the 4th term of the sequence.

New Price a After 1 year arAfter 2 years ar2 After 3 years

ar3 continued on next slide


Solution continued: Therefore ar3 = £11,054.25

Substitute in our value of r to get:a(0.85)3 = £11,054.25a = £11,054.25/(0.85)3 = £18,000

Therefore the new price of the car was £18,000.

For our geometric sequences we have a = 18,000 and r = 0.85

With these two values we can then find the value of n which gives a value less than 5000.

continued on next slide


Solution continued:

Solve for arn = 5000

18000(0.85)n = 5000 substitute in for a and r(0.85)n = 5000/18000(0.85)n = 5/18log (0.85)n = log (5/18) take log of both sidesn log (0.85) = log (5/18) bring power forwardn = log (5/18) / log (0.85) = 7.88 evaluate

Therefore the value of the car will fall below £5000 after 7.88 years.


Example 3:In the sequence, 2, 6, 18, 54, ... Which is the first

term to exceed 1 000 000?

Solution:From the given sequence a = 2 and r = 3.

If we want to find the first term to exceed 1 000 000 we solve arn = 1 000 000

Substitute in the values we have to get:2(3)n = 1 000 000(3)n = 500,000 dividing across by 2

continued on next slide


Solution continued:

(3)n = 500,000log (3)n = log (500,000) take log of both sidesn log (3) = log (500,000) bring power forwardn = log (500,000) / log (3) take numbers to one siden = 11.94 evaluate

Therefore the first term to exceed 1 000 000 is the 12th term as we cannot have an 11.94th term.


ASSIGNMENT:

This weeks assignment is in Moodle.

Please follow the link in the course area. Attempt all questions and include working out.

Deadline is 5:00pm on Monday 22nd March 2010.

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