### Balance Calculation:
To find the balance in the account after 30 years with an initial deposit of [tex]$1200 and an Annual Percentage Rate (APR) of 3.5%, compounded annually, we use the formula for compound interest:
\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]
where:
- \( A \) is the amount of money accumulated after \( n \) years, including interest.
- \( P \) is the principal amount (initial deposit), which is $[/tex]1200.
- [tex]\( r \)[/tex] is the annual interest rate (decimal), which is 0.035.
- [tex]\( n \)[/tex] is the number of times interest is compounded per year, which is 1 for annual compounding.
- [tex]\( t \)[/tex] is the time the money is invested for, which is 30 years.
For this problem:
- [tex]\( P = 1200 \)[/tex]
- [tex]\( r = 0.035 \)[/tex]
- [tex]\( n = 1 \)[/tex]
- [tex]\( t = 30 \)[/tex]
The formula simplifies to:
[tex]\[ A = 1200 \left(1 + 0.035\right)^{30} \][/tex]
By computing this, the balance [tex]\( A \)[/tex] will be:
[tex]\[ A \approx 3368.15 \][/tex]
So, the balance in the account after 30 years would be approximately [tex]$3368.15.
### Interest Earned:
The interest earned over the entire time period is the difference between the final balance and the initial deposit.
\[ \text{Total Interest} = A - P \]
From our previous calculation:
- \( A = 3368.15 \)
- \( P = 1200 \)
Thus,
\[ \text{Total Interest} = 3368.15 - 1200 \approx 2168.15 \]
So, the total interest earned over 30 years would be approximately $[/tex]2168.15.
