Question 1133531
Amount to contribute is A=m(1+(r/n))^360-1/(r/n)
Amount to withdraw requires m(1-(1+(r/n)^-(nt)/(r/n)
m is the amount withdrawn each month, so 10000(1-(1+.05/12)^(-240))/(.05/12)
=$1515253.13, rounding at the end.  

This is equal to m[(1+(r/n)^360)-1]/(r/n)
$1820.65 ANSWER