A=p[(1−i)n−1]/iA=p[(1-i)^n-1]/iA=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)A=3000[(1+0.04)A=3000[(1+0.04)22 −1]/0.04-1]/0.04−1]/0.04
A=75000[1.04A=75000[1.04A=75000[1.0422−1]/0.04-1]/0.04−1]/0.04
A=102743.90A=102743.90A=102743.90
Final amount is = $ 102743.90
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