Transcript Sullivan College Algebra Section 6.6

Sullivan PreCalculus
Section 4.7
Compound Interest
Objectives of this Section
• Determine the Future Value of a Lump Sum of Money
• Calculate Effective Rates of Return
• Determine the Present Value of a Lump Sum of Money
• Determine the Time Required to Double or Triple a Lump
Sum of Money
Interest is the money paid for the use of
money. The total amount borrowed is called
the principal. The rate of interest,
expressed as a percent, is the amount charged
for the use of the principal for a given period
of time, usually on a yearly (per annum)
basis.
Simple Interest Formula
I = Prt
The amount A after t years due to a principal
P invested at an annual interest rate r
compounded n times per year is
r

A  P1  
 n
nt
Suppose your bank pays 4% interest per
annum. If $500 is deposited, how much will
you have after 3 years if interest is
compounded …
a) Annually
nt
1( 3)
r
0.04


A  P 1    $500 1 
  $562.43
 n

1 
(b) Monthly
nt
12 ( 3)
r
0.04


A  P 1    $500 1 
  $563.64
 n

12 
The amount A after t years due to a principal P
invested at an annual interest rate r
compounded continuously is
A  Pe
rt
Suppose your bank pays 4% interest per
annum. If $500 is deposited, how much will
you have after 3 years if interest is
compounded continuously?
A  Pe  $500e
rt
0.04 ( 3)
 $563.75
The present value P of A dollars to be received
after t years, assuming a per annum interest
rate r compounded n times per year, is
r

P  A1  
 n
nt
If interest is compounded continuously, then
rt
P  Ae
How much should you deposit today in order
to have $20,000 in three years if you can earn
6% compounded monthly from a bank C.D.?
r

P  A 1  
 n
 nt
0.06

 $20,000 1 


12 
 $16,712.90
12 ( 3)
How long will it take to double an investment
earning 10% per annum compounded
quarterly?
r

A  P 1  
 n
nt
010
. 

ln 2  ln 1 


4 
010
. 

2 P  P 1 


4 
4t
4t
010
. 

ln 2  4t ln 1 


4 
ln 2
t
0.10

4 ln 1 


4 
 7.018 years