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A fund of 2,000 is to be accumulated by n annual payments of 50 by the end of each year, followed by n annual payments of 100 by the end of each year, plus a smaller final payment made 1 year after the last regular payment. If the effective rate of interest is 4.5%, find n and the amount of the final irregular payment.

Correct answer: $n=9$, payment$=32.42$.

From looking online, the correct solution is by initially solving for $n$ from $$ 50a_{n|.045}(1.045)^n + 100a_{n|.045} = 2000 $$

However, the question says 2000 is to be accumulated by the two annuities plus a final drop payment. So shouldn't the equation of value be

$$ 50a_{n|.045}(1.045)^{n+1} + 100a_{n|.045}(1.045) + X= 2000 $$

(I understand that from the above equation it's not possible to solve for $n$ because $X$ is also an unknown. But I don't understand how the wording of the question implies the first equation.)

MinYoung Kim
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1 Answers1

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What I would have thought you were looking for was the solution to

$$\sum\limits_{i=n+1}^{2n} 50\times 1.045^i + \sum\limits_{i=1}^{n} 100\times 1.045^i +X=2000$$ where you want $n$ to be an integer and $X$ non-negative but as small as possible

One approach could be to solve the equation without $X$ and then round the resulting $n$ down to an integer and then find $X$. So with $$\sum\limits_{i=n+1}^{2n} 50\times 1.045^i + \sum\limits_{i=1}^{n} 100\times 1.045^i +X=2000$$ i.e. using geometric series

$$50\left(1.045^{2n} -1.045^{n}\right) + 100\left(1.045^{n} -1\right) = \frac{2000\times 0.045}{1.045} $$

you can get

$$n = \dfrac{\log\left(\frac12\left(\sqrt{9 + \frac{2000 \times 0.045 \times 4}{1.045 \times 50}}-1\right)\right)}{\log(1.045)} \approx 9.107$$

Round this down to $n=9$ and then you can use the original equation to say $$X= 2000 -\tfrac{1.045}{0.045} \left(50\left(1.045^{18} -1.045^{9}\right) + 100\left(1.045^{9} -1\right)\right)\approx 32.41$$

Henry
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