Geometric Sequences & Series
This week the focus is on using geometric
sequences and series to solve problems
involving growth and decay.
Geometric Sequences & Series
CONTENTS
Growth and Decay
Interest Rates
Interest Rates Illustration
Example 1
Example 2
Example 3
Assignment
Geometric Sequences & Series
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.
Geometric Sequences & Series
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.
Geometric Sequences & Series
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
Geometric Sequences & Series
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 & Series
Solution 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 2
n log(1.035) = log 2
n = log 2/log 1.035 = 20.15
Therefore my money will have doubled after 20.15
years.
Geometric Sequences & Series
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
ar
After 2 years
ar2
After 3 years
ar3
continued on next slide
Geometric Sequences & Series
Solution continued:
Therefore ar3 = £11,054.25
Substitute in our value of r to get:
a(0.85)3 = £11,054.25
a = £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
Geometric Sequences & Series
Solution continued:
Solve for arn = 5000
18000(0.85)n = 5000
(0.85)n = 5000/18000
(0.85)n = 5/18
log (0.85)n = log (5/18)
n log (0.85) = log (5/18)
n = log (5/18) / log (0.85) = 7.88
substitute in for a and r
take log of both sides
bring power forward
evaluate
Therefore the value of the car will fall below £5000 after
7.88 years.
Geometric Sequences & Series
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
Geometric Sequences & Series
Solution continued:
(3)n = 500,000
log (3)n = log (500,000)
n log (3) = log (500,000)
n = log (500,000) / log (3)
n = 11.94
take log of both sides
bring power forward
take numbers to one side
evaluate
Therefore the first term to exceed 1 000 000 is the 12th
term as we cannot have an 11.94th term.
Geometric Sequences & Series
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.