Question

Deposits of 5.800$ are made every 6 months into a fund starting today and continuing until 10 deposits have been made. Six months after the last deposit there is 70.630$ in the fund. What nominal rate of interest, j2 do the deposits earn?

Answer 1

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

$70630=5800(\frac {(1+r)^{10}-1}{r}) \\ \frac{70630}{5800}=\frac {(1+r)^{10}-1}{r} \\ 12.177 \times r = (1+r)^{10} -1 \\ r ≈ 0.04304$

nominal rate of interest j2 $= 0.04304 \times 2 = 0.08608 = 8.6 \; \%$