A borrower has has the following options for repaying a loan:
- sixty monthly payments of $100$ at the end of each month
- A single payment of $6000$ at month $k$
Interest is a nominal rate of $.12$ per month meaning $i=.12/12=.01$.
Find $k$.
I know the first part is an annuity immediate with $$100 \frac{1-(1.01)^{-60}}{.01}=4495.504.$$
Now we want to find when $6000$ of payment will be equal to the $60$ monthly payments of $100$ which tells me want to know when will the accumulation of $6000$ at $.12%$ will be equal. Since we are working with present value $k$ will be negative
$$4495.504=6000(1.01)^{-k} \to 28.9.$$
If the assumption is that $k$ is chosen in such a way that both options result in the same present value, then the equation to be solved is $$6000 v^k = 100 a_{\overline{60}\rceil j} = 100 \frac{1 - v^{60}}{j},$$ where $j = 0.01$ is the monthly effective rate of interest and $v = 1/(1+j)$ is the monthly present value discount factor. Therefore, $$k = \frac{\log (1 - v^{60}) - \log 60 j}{\log v} \approx 29.012273332\ldots.$$ Your answer is off slightly, for reasons I cannot determine.
If rounded to the nearest integer, this corresponds to a single lump sum payment at the end of $29$ months, with present value $$6000 v^{29} \approx 4496.05,$$ whereas the annuity has present value $4495.50$, for a difference of approximately $0.55$. Consequently, the payment of $6000$ is a slight overpayment compared to the annuity: to get the same present value, this lump sum can be adjusted downward to $5999.2673\ldots$.