5
Copyright (c) 2003 Brooks/Cole, a division of Thomson Learning, Inc

Simple Interest - interest that is compounded on the original principal only.
Interest: I = Prt
Accumulated amount: A = P (1 + rt )
Copyright (c) 2003 Brooks/Cole, a division of Thomson Learning, Inc

Ex.
$800 is invested for 9 years in an account that pays 12% annual simple interest. How much interest is earned? What is the accumulated amount in the account?
P = $800, r = 12%, and t = 9 years
Interest: I = Prt
= (800)(0.12)(9)
= 864
Accumulated amount = principal + interest or $864
= 800 + 864 = 1664 or $1664
Copyright (c) 2003 Brooks/Cole, a division of Thomson Learning, Inc

Compound Interest – interest is added to the original principal and then earns interest at the same rate.
A

P
 i n
(1 ) where i
 r m
and n
 mt
A = Accumulated amount after n periods
P = Principal r = Nominal interest rate per year m = Number of conversion periods per year t = Term (number of years)
Copyright (c) 2003 Brooks/Cole, a division of Thomson Learning, Inc

Ex.
Find the accumulated amount A , if $4000 is invested at 3% for 6 years, compounded monthly.
P = $4000, r = 3%, t = 6, and m = 12
So i
 r m

.03
12

.0025 and n
 mt

12(6)

72
A

P
 i n
(1 )


4787.79
or $4787.79
Copyright (c) 2003 Brooks/Cole, a division of Thomson Learning, Inc

Effective Rate – the simple interest rate that would produce the same accumulated amount in 1 year as the nominal rate compounded m times per year. r eff
 
 r m  m

1 where r eff
= Effective rate of interest r = Nominal interest rate per year m = Number of conversion periods per year
Copyright (c) 2003 Brooks/Cole, a division of Thomson Learning, Inc

Ex.
Find the effective rate that corresponds to a nominal rate of 6% compounded quarterly.
r = 6% and m = 4 r eff


 r m


 m

1
 

.06
4 
4

1

.06136
So about 6.136% per year.
Copyright (c) 2003 Brooks/Cole, a division of Thomson Learning, Inc

Present Value (principal) – the amount required now to reach the desired future value.
P

A
 i
 n
(1 ) where i
 r m
and n
 mt
Copyright (c) 2003 Brooks/Cole, a division of Thomson Learning, Inc

Ex.
Jackson invested a sum of money 10 years ago in an account that paid interest at a rate of 8% compounded monthly. His investment has grown to
$5682.28. How much was his original investment?
A = $5682.28, r = 8%, t = 10, and m = 12 i
 r
 m
.08
and
12 n
 mt

12(10)

120

120
P
 
.08
12

2560.00 or $2560
Copyright (c) 2003 Brooks/Cole, a division of Thomson Learning, Inc

Annuity – a sequence of payments made at regular time intervals.
Ordinary Annuity – payments made at the end of each payment period.
Simple Annuity – payment period coincides with the interest conversion period.
Copyright (c) 2003 Brooks/Cole, a division of Thomson Learning, Inc

The future value S of an annuity of n payments of R dollars each, paid at the end of each investment period into an account that earns interest at the rate of i per period is
S R


(1
 i ) n 
1 i



Copyright (c) 2003 Brooks/Cole, a division of Thomson Learning, Inc

Ex.
Find the amount of an ordinary annuity of 36 monthly payments of $250 that earns interest at a rate of 9% per year compounded monthly.
R = 250, n = 36 and i

.09
12
S R


(1
 i ) n 
1 i



S

250




 
1
 .09
12
.09
12

 
36

1





S

10288.18 or $10,288.18
Copyright (c) 2003 Brooks/Cole, a division of Thomson Learning, Inc

The present value P of an annuity of n payments of R dollars each, paid at the end of each investment period into an account that earns interest at the rate of i per period is
P R


  i
 n
1 (1 ) i



Copyright (c) 2003 Brooks/Cole, a division of Thomson Learning, Inc

Ex.
Paige’s parents loaned her the money to buy a car. They required that she pay $150 per month, for
60 months, with interest charged at 2% per year compounded monthly on the unpaid balance. What was the original amount that Paige borrowed?
P R



150
  i )
 n i






1
  
 
 
60





8557.85
or $8557.85
Copyright (c) 2003 Brooks/Cole, a division of Thomson Learning, Inc

The periodic payment R on a loan of P dollars to be amortized over n periods with interest charged at a rate of i per period is
R

Pi
  i )
 n
Copyright (c) 2003 Brooks/Cole, a division of Thomson Learning, Inc

Ex.
The Kastners borrowed $83,000 from a credit union to finance the purchase of a house. The credit union charges interest at a rate of 7.75% per year on the unpaid balance, with interest computations made at the end of each month. The Kastners have agreed to repay the loan in equal monthly installments over
30 years. How much should each payment be if the loan is to be amortized at the end of the term?
P = 83000, n = (30)(12) = 360, and i

0.0775
12
Continued
Copyright (c) 2003 Brooks/Cole, a division of Thomson Learning, Inc

R

Pi
  i )
 n

1

83000
1


.0775
12
.0775
12


 
360

594.62
So a monthly installment of
$594.62
Copyright (c) 2003 Brooks/Cole, a division of Thomson Learning, Inc

Ex.
A bank has determined that the Radlers can afford monthly house payments of at most $750.
The bank charges interest at a rate of 8% per year on the unpaid balance, with interest computations made at the end of each month. If the loan is to be amortized in equal monthly installments over 15 years, what is the maximum amount that the Radlers can borrow from the bank?
R = 750, n = (15)(12) = 180, and i

0.08
12
Continued
Copyright (c) 2003 Brooks/Cole, a division of Thomson Learning, Inc

P

R


1
   i n



750



1
  
 
 
180





78480.44
So they can borrow up to about $78480.44
Copyright (c) 2003 Brooks/Cole, a division of Thomson Learning, Inc

The periodic payment R required to accumulate S dollars over n periods with interest charged at a rate of i per period is
R
 iS
(1
 i ) n 
1
Copyright (c) 2003 Brooks/Cole, a division of Thomson Learning, Inc

Ex.
Max has decided to set up a sinking fund for the purpose of purchasing a new car in 4 years.
He estimates that he will need $25,000. If the fund earns 8.5% interest per year compounded semi-annually, determine the size of each (equal) semi-annual installment that Max should pay into the fund.
S = 25000, n = 4(2) = 8, and i

0.085
4
Continued
Copyright (c) 2003 Brooks/Cole, a division of Thomson Learning, Inc

R
 iS
(1
 i ) n 
1


1

.085


4
.085

25000
4
 
8

1

2899.91
So semi-annual payments of about $2899.91
Copyright (c) 2003 Brooks/Cole, a division of Thomson Learning, Inc

Arithmetic progression – a sequence of numbers in which each term after the first is obtained by adding a constant d (common difference) to the preceding term.
Ex.
1, 8, 15, 22, 29, …
First term Common difference: d = 7
Copyright (c) 2003 Brooks/Cole, a division of Thomson Learning, Inc

The n th term with the first term a and common difference d is given by a n

1
 d
The sum of the first n terms with the first term a and common difference d is given by
S n
 n
2

2 a
 
1) d

Copyright (c) 2003 Brooks/Cole, a division of Thomson Learning, Inc

Ex.
Given the arithmetic progression
1, 8, 15, 22, 29, … find the 10 th term and the sum of the first 10 terms.
10 th term:
Sum: a a = 1, d = 7, and n = 10.
n

1
 d a
10
     
= 64
S n
 n
2

2 a
 
1) d

S
10

10 
2

= 325
Copyright (c) 2003 Brooks/Cole, a division of Thomson Learning, Inc

Geometric progression – a sequence of numbers in which each term after the first is obtained by multiplying the preceding term by constant r
(common ratio).
Ex.
9, 3, 1, 1/3,…
Common ratio: r = 1/3
Copyright (c) 2003 Brooks/Cole, a division of Thomson Learning, Inc

The n th term with the first term a and common ratio r is given by a n
 ar n

1
The sum of the first n terms with the first term a and common ratio r is given by
S n


 a na

1
1

 r r n

if
if r r


1
1
Copyright (c) 2003 Brooks/Cole, a division of Thomson Learning, Inc

Ex.
Given the geometric progression
4, 12, 36, 108, … find the 8 th term and the sum of the first 8 terms.
a = 4, r = 3, and n = 8.
8 th term:
Sum: S n a n
 ar a
8


1 n
 
4 3 = 8748 a

1
 r n



1
 r
 8

S
8

= 13120
Copyright (c) 2003 Brooks/Cole, a division of Thomson Learning, Inc