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Interest Over Several Periods


In this section, you’ll learn what happens when n has a value greater than 1. In other words, there is more than one time period over which interest will compound.

Over time, this can lead to significant increases in the balance, because compounding interest means that interest is being calculated based on prior interest.

When saving, compound interest is quite appealing, because it gives you more money. But with credit cards and loans, compound interest means you end up paying interest charges on interest you’ve already been charged!

Formula

Interest calculations

Kn = K0 (1 + p 100) n

where K0 is the initial balance, and p is the interest rate. After n periods (or number of times the interest is added) the capital is Kn.

Example 1

Interest over time

You borrow $5000 from the bank to finance your graduation party. The bank decides that you have to pay 40% in interest per year. You ask to pay the total amount in five years. How much do you have to pay back?

The initial balance is K0 = 5000. Kn is unknown, p = 40 and n = 5. You can put these figures straight into the formula and solve the equation: Kn = 5000 (1 + 40 100) 5 = 5000 1.45 = 26891.2

You will owe $26891.2 to the bank after 5 years!

Example 2

Interest over time

Your aunt deposited money in the bank 20 years ago. Interest has been accumulating at a rate of 4.5% per year. Today, there is $25000 in the account. How much was originally deposited?

In this example, the initial balance K0 is the unknown variable. Kn = 25000, p = 4.5 and n = 20. Just plug in the figures you do know into the formula and solve the equation:

Kn = K0 (1 + p 100) n 25000 = K0 (1 + 4, 5 100) 20 25000 = K0 1.04520 25000 1.04520 = K0 10366.07 K0

Kn = K0 (1 + p 100) n 25000 = K0 (1 + 4, 5 100) 20 25000 = K0 1.04520| ÷ 1.04520 10366.07 K0

Your aunt deposited around $10366 20 years ago.

Example 3

Interest over time

Your grandfather was always a cool guy. In his will, he has left his classic car to you. He bought the car 55 years ago, and it cost $5000 at the time. The annual increase in the car’s value is 5%. How much is the car worth today?

Again, begin by finding out where in the formula the different numbers fit. In this case, K0 = 5000, p = 5, n = 55, and Kn is the unknown. Plug these straight into the formula and solve the equation to find the answer: Kn = K0 (1 + p 100) n = 5000 (1 + 5 100) 55 = 5000(1 + 0.05)55 = 5000 1.0555 73178

The car is worth $73178 today.

Example 4

Find the interest rate

Ola bought a car for $25000 15 years ago. The present value of the car is $2200. What was the yearly percentage decrease in value for Ola’s car?

Again you can begin by finding where the numbers go in the formula. In this case we know that K0 = 25000, Kn = 2200, n = 15, and p is the unknown we want to find. Insert this information into the formula to get the answer. As p is unknown inside the parenthesis raised to the power of 15, you need to take the 15th root to solve the equation:

25000 = 2200 (1 p 100) 15 25000 2200 = (1 p 100) 15 11.36 = (1 p 100) 15 0.08815 = (1 p 100) 1515 0.85 = 1 p 100 0.15 = p 100 p = 15

25000 = 2200 (1 p 100) 15 | ÷ 22000 11.36 = (1 p 100) 15 0.08815 = (1 p 100) 1515 0.85 = 1 p 100 0.15 = p 100 | 100 p = 15

The yearly decrease in value was 15 %.

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Interest Over One Period
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