Connor has made deposits of $\mathrm{\$}85.00$ into his savings account at the end of every three months for 14 years. If interest is $11\mathrm{\%}$ per annum compounded monthly and he leaves the accumulated balance for another 4 years, what would be the balance in his account then?
The balance in his account would be $\mathrm{\$}◻$.
(Round the final answer to the nearest cent as needed. Round all intermediate values to six decimal places as needed.)

Step 1 ：Let's denote the total amount in the account after 14 years as $A1$ and the total amount in the account after another 4 years as $A2$.

Step 2 ：Since Connor deposits money every three months, there are 4 deposits in a year. So, in 14 years, there are $14\ast 4=56$ deposits.

Step 3 ：The interest is compounded monthly, so the interest rate per period is $\frac{11\mathrm{\%}}{12}=0.00916667$.

Step 4 ：We can use the formula for the future value of an ordinary annuity to calculate $A1$: $A1=P\ast [(1+r{)}^{nt}-1]/r$, where $P$ is the amount of each deposit, $r$ is the interest rate per period, and $nt$ is the total number of periods.

Step 5 ：Substituting the given values, we get: $A1=85\ast [(1+0.00916667{)}^{56}-1]/0.00916667=\mathrm{\$}7,238.96$.

Step 6 ：Next, we need to calculate $A2$. The amount $A1$ will be compounded for another 4 years at the monthly interest rate of 0.00916667.

Step 7 ：We can use the formula for compound interest to calculate $A2$: $A2=A1\ast (1+r{)}^{nt}$, where $A1$ is the initial amount, $r$ is the interest rate per period, and $nt$ is the total number of periods.

Step 8 ：Substituting the given values, we get: $A2=7238.96\ast (1+0.00916667{)}^{(4\ast 12)}=\mathrm{\$}10,315.68$.

Step 9 ：So, the balance in his account would be $\boxed{{\displaystyle \mathrm{\$}10,315.68}}$.