Future value of the annual deposit
"= contribution \\times (\\frac {(1+r)^{p}-1}{r})"
r is rate of interest
p is no. of period
nominal rate of interest j2 = r*2
p=10
contribution=5800
final value =70630
"70630=5800(\\frac {(1+r)^{10}-1}{r}) \\\\\n\n \\frac{70630}{5800}=\\frac {(1+r)^{10}-1}{r} \\\\\n\n12.177 \\times r = (1+r)^{10} -1 \\\\\n\nr \u2248 0.04304"
nominal rate of interest j2 "= 0.04304 \\times 2 = 0.08608 = 8.6 \\; \\%"
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