## Subset

Subset

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

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

## 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

## Future Value Of an Ordinary Annuity

Future Value Of an Ordinary Annuity

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

## 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 = or P = .

## Effective Rate

Effective Rate

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

re  =

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   =

= .050115577 = 5.01 %

## Compound Amount

Compound Amount

A = P( 1 + i)n,

Where i = 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 = = .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 = of a year. The maturity value is

A = P( 1 + rt)

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 = (in years). The total interest she will pay is

I = 366.67.