Archive for July, 2014

Subset

Subset

Set A is a subset of set B (written A \subseteq b if every element of A is also an element

of B. Set A is a proper subset (written A\subset B if A \subseteq b and A \subseteq B.

 

 

Present value of an Annuity

Present value of an Annuity

 

The present value P of an annuity of n payments of R dollars each at the end of

consecutive interest periods with interest compounded at a rate of interest i per

period is

 

P=R[ \frac{1-(1+i)^{-n}}{i}]

Future Value Of an Ordinary Annuity

Future Value Of an Ordinary Annuity

S=R[\frac{(1+i)^{n}-1}{i}]

 

Where

S is the future value;

R is the payment;

i is the interest rate per period;

n is the number of periods.

 

Sum Of Terms

Sum Of Terms

 

If a geometric sequence has first term a and common ratio r, then the sum Sn of the first n terms is given by

 

S_{n}=\frac{a(r^{n}-1)}{r-1},r\neq 1.

Present Value for Compound Interest

Present Value for Compound Interest

 

The present value of A dollars compounded at an interest rate I per period for n periods is

P = \frac{A}{(1+i)^{n}} or P = A(1+i)^{-n}.

Effective Rate

Effective Rate

The effective rate corresponding to a stated rate of interest r compounded m times per years is

re  =   (1 + \frac{r}{m})^{m} - 1.

Example:-

A bank pays interest of 4.9% compounded monthly. Find the effective rate.

Solution:-

Use the formula given above with r = .049 and m = 12. The effective rate is

re   =   (1 + \frac{.049}{12})^{12} - 1.

= .050115577 = 5.01 %

Compound Amount

Compound Amount

A = P( 1 + i)n,

Where i = \frac{r}{m} and n = mt,

A is the future (maturity) value;

P is the principal;

 r is annual interest rate;

m is the number of compounding periods per year;

t is the number of years;

n is the number of compounding periods;

i is the interest rate per period.

Example:- Suppose 1000 is deposited for 6 years in an account paying 4.25% per year compounded annually.

 

(a)     Find the compound amount.

Solution:- In the formula above, P = 1000, I = .0425, and  n = 6(1) = 6.

A = P(1 + i)n

A = 1000(1.0425)6

Using a calculator, we get

 

A = 1283.68 the compound amount.

 

(b)     Find the amount of interest earned.

 

Soluion:-  Subtract the initial deposit from the compound amount.

Amount of interest = 1283.68 – 1000 = 283.68

 

 

Simple Interest Rate Find

 Simple  Interest Rate Find

 

Example:-

Carter Fenton wants to borrow 8000 from Christine O’Brien. He is willing to pay back 8380 in 6 months. What interest rate will he pay?

 

Solution:-

Use the formula for future value, with A = 8380,   P = 8000, t = \frac{6}{12} = .5, and solve for r.

A = P(1 + rt)

8380 = 8000(1 + .5r)

8380 = 8000 + 4000r (Distributive property)

380 = 4000r (Subtract 8000)

r = .095

Thus, the interest rate is 9.5%

 

Future or Maturity Value for Simple Interest

Future or Maturity Value for Simple Interest

The future or maturity value A of P dollars at a simple interest rate r for t years is

A = P(1 + rt).

Example:-

A loan of  2500 to be repaid in 8 months with interest of 9.2%

Solution:-

The loan is for 8 months, or \frac{8}{12} = \frac{2}{3} of a year. The maturity value is

A = P( 1 + rt)

A = 2500[1 + .092(\frac{2}{3})]

A = 2500(1 + .06133) = 2653.33

or  2653.33. (The answer is rounded to the nearest cent, as is customary in financial problems.) Of this maturity value.

2653.33 –  2500 =  153.33

Represents interest.

 

Simple Interest

Simple Interest

I = Prt,

where  p is the principal;

r is the annual interest rate;

t us the time in years.

Example:-

To bye furniture for a  new apartment, Jennifer Wall borrowed 5000 at 8% simple interest for 11 months. How much interest will she pay ?

Solution:-

From the formula, I = Prt, with P = 5000, r = .08, and t = \frac{11}{12}(in years). The total interest she will pay is

I = 5000(.80)(\frac{11}{12}) = 366.67

I = 366.67.