Asked by Jennifer Aguilar on May 06, 2024

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A deposit of $3,000\$ 3,000$3,000 is made in an account that earns 4%4 \%4% interest compounded yearly. The balance in the account after N years is given by AN=3,000(1+0.04) NA _ { N } = 3,000 ( 1 + 0.04 ) ^ { N }AN=3,000(1+0.04) N , N=1,2,3,…N = 1,2,3 , \ldotsN=1,2,3, Find the balance in this account after 202020 years by computing A20A _ { 20 }A20 . Round your answer to the nearest cent.

A) A20=$5,403A _ { 20 } = \$ 5,403A20=$5,403
B) A20=$62,400A _ { 20 } = \$ 62,400A20=$62,400
C) A20=$6,573A _ { 20 } = \$ 6,573A20=$6,573
D) A20=$3,245A _ { 20 } = \$ 3,245A20=$3,245
E) A20=$6,321A _ { 20 } = \$ 6,321A20=$6,321

Compounded Yearly

Calculating interest on an investment or loan annually, where the accrued interest is added to the principal sum each year.

Interest Rate

The proportion of a loan charged as interest to the borrower, typically expressed as an annual percentage of the outstanding loan.

Balance

A condition in which different elements are equal or in the correct proportions, often used in financial contexts to denote the amount of money available in an account.

  • Apply principles of sequences and series to real-world problems.
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ZK
Zybrea KnightMay 09, 2024
Final Answer :
C
Explanation :
Using the formula $A_N = P(1+r)^N$, where $P$ is the principal deposit, $r$ is the interest rate as a decimal, and $N$ is the number of years:

$A_{20} = 3000(1+0.04)^{20} \approx 6573.15$

Rounding to the nearest cent, the answer is $\boxed{\textbf{(C) } A_{20} = \$6,573}$.