Question

l. On 1 April 2018, €1,250 was invested in EUROBANK at a fixed rate of 4.5% per withdrawals are to be made. Find: (a) a(1), the amount in the account after 1 year - i.e., on 1 April 2019; [2 marks] [2 marks] 5 marks) 5 marks) (b) a(2), the amount in the account after 2 years-ie, on 1 April 2020; c) A recursive formula for a(n), the amount in the account after n years (d) A closed form solution for a(n), the amount in the account after n years. (e) The amount in the account on 1 April 2061.

Answer 1

The formula for the amount (n) years after the investment is done compounded annually =

A = P * [ 1 + { R / 100 } ] ^{^ t} , where

A is the amount after (n) years of investment.

P is the principal amount invested = $\euro$ 1250

R is the fixed rate of interest = 4.5%

t is the time period of investment in years.

(a) The amount in the account after 1 year ,i.e., on 1 April 2019

A₂₀₁₉ = 1250 * [ 1 + ( 4.5 / 100 ) ] ^{^ 1}

= 1250 * [ 1 + 0.045 ]

= 1250 * 1.045

A₂₀₁₉ = 1306.25

(b) The amount in the account after 2 years ,i.e., on 1 April 2020

A₂₀₂₀ = 1250 * [ 1 + ( 4.5 / 100 ) ] ^{^ 2}

= 1250 * [ 1 + 0.045 ] ^{^ 2}

= 1250 * (1.045) ^{^ 2}

= 1250 * 1.092025

A₂₀₂₀ = 1365.03125

(d) Closed form solution for amount in the account after (n) years is

A = P * [ 1 + { R / 100 } ] ^{^ t}, where

A is the amount after (n) years of investment.

P is the principal amount invested = $\euro$ 1250

R is the fixed rate of interest = 4.5%

t is the time period of investment in years.

(e) Amount in the account on 1 April 2061.

No. of years of investment (t) = 2061 - 2018

= 43 years.

Therefore,

A₂₀₆₁ = 1250 * [ 1 + ( 4.5 / 100 ) ] ^{^ 43}

= 1250 * [ 1 + 0.045 ] ^{^ 43}

= 1250 * (1.045) ^{^ 43}

= 1250 * 6.63744

A₂₀₆₁ = 8296.797726

(c) Recursive formula for amount in the account after (n) years.

I am not aware of the recursive formula for calculating the amount after (n) years of investment at compounding annually. I am aware of only 1 formula for calculating the amount, which I have mentioned in the solution.