1. Compound Interest I.. Compound Interest: A = P ( 1 + r / n ) nt A = Account balance after time has passed. P = Principal: $ you put in the bank. r = interest rate (written as a decimal). n = number of times a year the interest is compounded. (annual = 1, semi-annual = 2, quarterly = 4, monthly = 12, etc.) t = time (in years) the money is in the bank. A) To determine the Account balance after time has passed, plug all the #s into the formula and simplify.

2. Compound Interest Examples: 1) If you deposit $4000 in an account that pays 2.92% interest semi-annually, what is the balance after 5 years? How much did the account earn in interest? A = P ( 1 + r / n ) nt  A = 4000 ( 1 +.0292 / 2 ) 25 A = 4000 ( 1 +.0146 ) 10 A = 4000 (1.0146) 10 A = $ 4623.90 So the account gained $623.90 dollars in the 5 years.

3. Compound Interest Examples: 2) If you deposit $12,500 in an account that pays 4.5% interest quarterly, what is the balance after 8 years? How much did the account earn in interest? A = P ( 1 + r / n ) nt  A = 12500 ( 1 +.045 / 4 ) 48 A = 12500 ( 1 +.01125 ) 32 A = 12500 (1.01125) 32 A = $ 17,880.64 So the account gained $5380.64 dollars in the 8 years.

4. Compound Interest II.. Solving for P in Compound Interest: A = P (1 + r / n ) nt A) Plug all the #s into the formula. B) Simplify the ( ) nt part. C) Divide both sides by the ( ) nt part.

5. Compound Interest Examples: 3) How much would you have to deposit in a savings CD paying 4.9% annually so that you will have $60,000 in your account after 12 years? A = P ( 1 + r / n ) nt  60,000 = P ( 1 +.049 / 1 ) 112 60,000 = P ( 1 +.049 ) 12 60,000 = P (1.049) 12 (1.049) 12 (1.049) 12 P = $ 33,795.20

6. Compound Interest III.. Solving for r in Compound Interest: A = P (1 + r / n ) nt A) Plug all the #s into the formula. B) Divide by the P part to get (1 + r / n ) nt by itself. C) Get rid of the exponent with a radical. 1) Use a reciprocal fractional exponent. D) Evaluate the A/P^( 1 / nt ) term with a calculator. E) Solve for r. 1) Move the decimal 2 places to get a %.

7. Compound Interest Examples: 4) Your Great Grandpa bought his bride to be an engagement ring valued at $200.00 back in 1928. It was appraised in 2008 as being worth $12,500.00. What was the rate of increase per year? A = P ( 1 + r / n ) nt  12500 = 200 ( 1 + r / 1 ) 180 12500 = 200 ( 1 + r ) 80 200 200 62.5 = (1 + r ) 80 62.5^( 1 / 80 ) = ( 1 + r ) 80(1/80)  1.053 = 1 + r (subtract the 1).053 = r so r = 5.3%