Question about interest rates

Question

Can someone help me with this

Anna borrows $50, 000$ to the bank at a nominal interest rate $i(12) = 6\%$, (compounded monthly). She repays this loan by doing monthly payments (at the end of each month) during $6$ years. If each of the $36$ rst payments are of $R$ and each one of the last $36$ payments are of $(R + 1000)$. Using geometric series, determine $R$, and the interest amount that Anna paid to the bank.

Answer 1

$50,000 = R \sum_\limits {n=1}^{36} \frac {1}{1.005^n} + (R+1000) \sum_\limits {n=37}^{72} \frac {1}{1.005^n}\\ 50,000= R \frac {1.005^{36} - 1}{(0.005) (1.005^{36})} + \frac {(R+1000)}{1.005^{36}}\frac {1.005^{36} - 1}{(0.005) (1.005^{36})} $

$50,000 \frac {(0.005)(1.005^{72})}{1.005^{36} - 1}= R (1.005^{36} + 1) + 1000\\ R =50,000 \frac {(0.005)(1.005^{72})}{1.005^{72} - 1} - \frac {1000}{1.005^{36} + 1}$

Total payments $= 72 R + 36,000$

Total Interest = Total payments - 50,000

Answer 2

The basic idea is that the cash flow can be regarded as an annuity-immediate of $R$ per month for $72$ months, plus a second annuity-immediate of $1000$ per month for the last $36$ months of payment. That is to say, the present value of the cash flow can be written as $$50000 = R a_{\overline{72}\rceil j} + 1000 v^{36} a_{\overline{36}\rceil j},$$ where $j = i^{(12)}/12 = 0.06/12 = 0.005$ is the effective monthly interest rate, and $v = 1/(1+j) \approx 0.995025$ is the effective monthly discount factor. Since $$a_{\overline{n}\rceil j} = \frac{1 - v^n}{j} = \frac{1 - (1+j)^{-n}}{j},$$ the computation is straightforward: $$R = 1000 \left( \frac{50 - v^{36} a_{\overline{36}\rceil j}}{a_{\overline{72}\rceil j}} \right) = 1000 \left( \frac{50 - (0.835645)(32.871)}{60.3395} \right) \approx 373.412.$$ The total amount paid is simply $72R + 36000$, and the total interest paid is $72R + 36000 - 50000$.

It is worth noting that one can write $$\frac{50 - v^{36} a_{\overline{36}\rceil j}}{a_{\overline{72}\rceil j}} = \frac{50j - (v^{36} - v^{72})}{1 - v^{72}} = \frac{0.25 - y + y^2}{1 - y^2},$$ where $y = v^{36} = 0.835645$, which simplifies the computation slightly since now the value of $R$ is expressible in terms of the single value $y$.

Answer 3

This is an old question but what the heck,

The residual debt after first 36 payments is: $L(1+i)^{36}-R\frac{(1+i)^{36}-1}{i}$

After the last 36 payments the residual debt is zero:

$L(1+i)^{72}-R\frac{(1+i)^{36}-1}{i}(1+i)^{36}-(R+1000)\frac{(1+i)^{36}-1}{i}=0$

$R=Li\frac{(1+i)^{72}}{(1+i)^{72}-1}-\frac{1000}{(1+i)^{36}+1}, L=50000, i=0.005$

R=373.42