If $i=0.09$, find $n$ and the amount of final payment.

A fund of $\$500 $is to be accumulated by$ n $annual payments of$ \$100$, plus a final payment as small as possible made one year after the last regular payment. If $i = 0.09$, find $n$ and the amount of final payment.

I have gotten as far as:

$$500 = 100 \times (1.09)^{n} + P(1.09)^{n + 1},$$ $$\frac{500 - 100(1.09)^{n}}{(1.09)^{n + 1}} = P.$$

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You pay $100$. After a year, that has grown to $109$, and you pay another $100$, making a balance of $209$. After another year, that $209$ has grown to ... how much? And you pay another $100$, making a balance of ... how much? And you do that one more year, and how much do you have? And what happens then?

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Let $x_k$ be the amount at year $k$, with $x_0 = 100$. Then we have $x_{k+1} = (1+r) x_k+100$, with $r=0.09$. A few terms suggests a general solution, $x_0 = 100, x_1 = 100(1+r)$, $x_2 = 100(1+r)^2+100(1+r),..., x_{n-1} = 100(1+r)^{n-1}+...+100$.

After the $n$th payment another year elapses before a final payment of $F_n$ is made, resulting in a total of $500$, ie, \begin{eqnarray} 500 &=& 100(1+r)^{n}+...+100(1+r) + F_n \\ & = & 100(1+r)\frac{1-(1+r)^n}{1-(1+r)} +F_n \\ & = & 100(1+\frac{1}{r})((1+r)^n-1) + F_n \end{eqnarray} Rearranging gives $F_n = 500-100(1+\frac{1}{r})((1+r)^n-1)$. Note that $F_n$ is decreasing from $F_0 = 500$, so we can compute $\max \{ n | F_n \geq 0 \}$ to find the answer.

(As a sanity check, with $r=0$ we would have $n=5$ and $F_5 = 0$, so we expect $n < 5$.)

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The future value of an annuity is the accumulated amount, including payments and interest, of a stream of payments made to an interest-bearing account.

• If the payments are made at the end of the time periods, so that interest is accumulated before the payment, the annuity is called an annuity-immediate. For an annuity-immediate, it is the value immediately after the $n$-th payment. The future value is given by: $$s_{\overline{n\,}\!|i} = \frac{(1+i)^n-1}{i}$$ where $n$ is the number of terms and $i$ is the per period interest rate. Future value is $$FV_{\text{imm.}}(i,n,R) = R \times s_{\overline{n}|i}$$ where $R$ is the recurrent payment.
• An annuity-due is an annuity whose payments are made at the beginning of each period. The future values of an annuity-due can be calculated through the formula: $$\ddot{s}_{\overline{n|}i} = (1+i) \times s_{\overline{n|}i} = \frac{(1+i)^n-1}{d}$$ where $d$ is the effective rate of discount given by $d=\frac{i}{i+1}$. And for a stream of payments $R$ $$FV_{\text{due}}(i,n,R) = R \times \ddot s_{\overline{n}|i}$$

In your problem we have $R=100,\, i=9\%, FV=500$ and we have to find the number $n$ of payments and the final payment $P$ (as small as possible). So we have to solve the equation $$FV=Rs_{\overline{n}|i}+P\quad\Longrightarrow 500=100 s_{\overline{n}|9\%}+P\tag 1$$ if the the payments are made at the end of the time periods, or $$FV=R\ddot s_{\overline{n}|i}+P\quad\Longrightarrow 500=100 \ddot s_{\overline{n}|9\%}+P\tag 2$$ if the the payments are made at the beginning of the time periods. In both case $n$ cannot be equal to 5 because $FV/R=500/100=5$ only if $i=0$. So we have $n<5$, and the minimum value we can choice is $n=4$. For $n=4$ we find

1. $s_{\overline{4}|9\%}=4.5731$ and $P=500-100\times 4.5731=42.69$
2. $\ddot s_{\overline{4}|9\%}=4.9847$ and $P=500-100\times 4.9847=1.52$

Thus, $n=4$ and the smallest final payment is $P=1.52$ for payment made at the beginning of the time periods (annuity-due).

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