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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.

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  • $\begingroup$ What did you try? $\endgroup$ – José Carlos Santos Sep 21 '17 at 17:57
  • $\begingroup$ I found that at after the 72 months, Anna will have paid 71 602,2139$ and I am stuck here .. $\endgroup$ – user483182 Sep 21 '17 at 18:04
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$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

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  • $\begingroup$ Where is the 1.01 coming from ?? $\endgroup$ – user483182 Sep 21 '17 at 18:26
  • $\begingroup$ Thanks a lot for the answer $\endgroup$ – user483182 Sep 21 '17 at 18:26
  • $\begingroup$ sorry, 1.005... I had 12% interest in my mind as I was typing $\endgroup$ – Doug M Sep 21 '17 at 18:26
  • $\begingroup$ In the first equation, I thought that 71 602,2139 should be equal to the summation of the payment ? $\endgroup$ – user483182 Sep 21 '17 at 18:42
  • $\begingroup$ I get 373.41 for the first 36 payments, 1,373.41 for the next 36, Total payments 62,886. $\endgroup$ – Doug M Sep 21 '17 at 19:31
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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 $y$.

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