Accumulated money at 2 august 2002 (one month before withdrawl)
=100(F/A,0.5%,12*6)(F/P,0.5%,(12*18)-1)
= 100(F/A,0.5%,72)(F/P,0.5%,215)
= 100 (86.40886)(2.92216)
=25250
Now, withdrawl of 1000 starting 2 september 2002
25250=1000(P/A,0.5%,n)
formula of (P/A,i,n)
putting values in above formula (0.5%=0.005)
solving for n
n=27.06 months or 27.06/12=2.25 years
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