Question

An investor makes three deposits into a fund, at the end of 1, 3, and 5 years. The amount of the deposit at time t is 100(1.025)t . Find the size of the fund at the end of 7 years, if the nominal rate of discount convertible quarterly is 4 41 .

Answer 1

Answer:

The size of the fund at the end of 7 years is $483.110

Explanation:

Number of quarters = 4

We are given that the nominal rate of discount convertible quarterly is 4/41

Discount rate in each quarter =$\frac{\frac{4}{41}}{4} = \frac{1}{41}.$

Let A is the value after discount and X is the original value:

$A = X - X(\frac{1}{41}) \\A=X(1 - \frac{1}{41}) \\A=\frac{40}{41}X\\X = \frac{41}{40}A$

Now To calculate the value after 7 years we need to multiply each value by the interest raised to the correct power.

$A=100 \times 1.025^1 \times \frac{41}{40}^{(7-1) \times 4}+100 \times 1.025^3 \times \frac{41}{40}^{(7-3) \times 4}+100 \times 1.025^5 \times \frac{41}{40}^{(7-5) \times 4}$

A=483.110

Hence  the size of the fund at the end of 7 years is $483.110