"A=p[(1-i)^n-1]\/i"
Where
A= is the final amount
p= the periodic deposit (3000)
i= is the periodic interest 8/(2x100) = 0.04
n= the number of periods = 11x2=22
"A=3000[(1+0.04)"22 "-1]\/0.04"
"A=75000[1.04"22"-1]\/0.04"
"A=102743.90"
Final amount is = $ 102743.90
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