Connor has made deposits of $\$ 85.00$ into his savings account at the end of every three months for 14 years. If interest is $11 \%$ 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 $\$ \square$. (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 \(14*4 = 56\) deposits.

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

Step 4 ：We can use the formula for the future value of an ordinary annuity to calculate \(A1\): \(A1 = P * [(1 + r)^{nt} - 1] / r\), 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: \(A1 = 85 * [(1 + 0.00916667)^{56} - 1] / 0.00916667 = \$7,238.96\).

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 \(A2\): \(A2 = A1 * (1 + r)^{nt}\), 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: \(A2 = 7238.96 * (1 + 0.00916667)^{(4*12)} = \$10,315.68\).

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