Question

Ted invests $1077 in a savings account with a fixed annual interest rate of 9% compounded three times per year. What will be the account balance after ten years? Please HELP! Im timed on it!

Answer 1

Answer: $2,614.16

Step-by-step explanation:

Initially, he has $1077.

We know that this has a fixed annual interest rate of 9% compounded three times per year.

(remember that 9% is 0.09 in decimal form)

Then in one year, this interest is compounded in 3 times.

So we actually apply 0.09/3 3 times.

So if in year zero, Ted has $1077.

after 1 year, Ted will have:

M = $1077*(1 + 0.09/3)^3

Where the power of 3 is because this interest rate is applied 3 times.

Now, after another year Ted will have:

M = $1077*(1 + 0.09/3)^3*(1 + 0.09/3)^3 = $1077*(1 + 0.09/3)^(2*3)

We already can see the pattern here, after N years,

M(N) = $1077*(1 + 0.09/3)^(3*N)

The account balance after 10 years can be computed by evaluating the above equation in N = 10.

M(10) = $1077*(1 + 0.09/3)^(3*10) = $2,614.16