Find the present value of a series of investment of $1 at the beginning of each year for 15 years with effective interest rate of 13.5% p.a.
Solution:
Given, p=$1,r=13.5% p.a.=0.135,n=15 yearsp=\$1,r=13.5\%\ p.a.=0.135, n=15\ yearsp=$1,r=13.5% p.a.=0.135,n=15 years
So, present value,PV=p[1−(1+r)−nr]PV=p[\dfrac{1-(1+r)^{-n}}{r}]PV=p[r1−(1+r)−n]
=1[1−(1+0.135)−150.135]=$6.298926≈$6.3=1[\dfrac{1-(1+0.135)^{-15}}{0.135}]=\$6.298926\approx\$6.3=1[0.1351−(1+0.135)−15]=$6.298926≈$6.3
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