Problem 66 - College Algebra

Question

A deposit of \(\$ 10,000\) is made in an account that earns \(8 \%\) interest compounded quarterly. The balance in the account after \(n\) quarters is given by the sequence $$a_{n}=10,000\left(1+\frac{0.08}{4}\right)^{n}, \quad n=1,2,3, \ldots$$ Find the balance in the account after six years. Round to the nearest cent.

Step-by-Step Solution

Answer

To find the balance in the account after six years, substitute 24 (the total quarters in six years) into the sequence formula and perform the calculation. Round off the answer to the nearest cent.

Step 1: Identify Relevant Information

Based on the question, the principal amount is \(\$10,000\), the interest rate is \(8\%\) (or \(0.08\) in decimal form), and the interest is compounded quarterly. The number of years is 6, but it needs to be converted to quarters since the compounding period is quarterly. A year has 4 quarters, so six years would be \(6*4 = 24\) quarters.

Step 2: Substitute into Sequence Formula

Having established that the total number of quarters \(n\) is 24, substitute this into the sequence formula \( a_{n} = 10,000\left(1+ \frac{0.08}{4}\right)^{n}\). This results in \( a_{24} = 10,000\left(1+ \frac{0.08}{4}\right)^{24}\).

Step 3: Compute the Balance

Perform the calculations inside the parentheses first due to mathematical order of operations (Remember BIDMAS/BODMAS rule - Brackets, Indices, Division and Multiplication, and Addition and Subtraction). Then raise the result to the power of 24, and finally multiply by 10,000. After computation, round the result to two decimal places to get the balance amount to the nearest cent.