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I'm struggling with the following questions.

A sum of £1000 is invested in an interest bearing bank account. Determine the accumulation of the account over 5 years and 4 months if:

$d^{3} = 6.5$% per annum.

where d represents the rate of discount to be paid 3 times per year. This is my solution

Accumulated Amount = $1000 * (1- \dfrac{d^{3}}{3})^{16} = 1000 * (1- \dfrac{0.065}{3})^{16} = 704.35$

I used 16 because that's how many 4 months period there in 5 years and 4 months and since the rate of discount is given as 3 times per year it means it's paid every 4 months.

Could somebody tell me if my think is correct ?

alpha787
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1 Answers1

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$d^{(3)}=6.5\% $, $n=5\times 3+1=16$, so we have $A(0)=1,000$

The amount after $n$ periods will be $$ A(n)=\frac{A(0)}{\left(1-\frac{d^{(3)}}{3}\right)^n} $$ and then $$ A(16)=\frac{1,000}{\left(1-\frac{0.065}{3}\right)^{16}}=1,419.75 $$

alexjo
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