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A $60$-month loan is too be repaid with level payments of $1000$ at the end of each month. The interest in the last payment is $7.44$. Calculate the total interest paid over the life of the loan.

Let effective interest be $j$.

I tried finding the outstanding loan balance after 59 payments using the prospective method.

$B_{59|j}^p=1000[\frac{1-(1+j)^{-59}}{j}]$

Interest = $7.44$

Letting L=loan amount and P=Payment

$7.44=L-1000\frac{1-(1+j)^{-59}}{j}$

$L=Pa_{60|j}=1000[\frac{1-(1+j)^-60}{j}]$

$7.44=1000[\frac{1-(1+j)^-60}{j}]-1000\frac{1-(1+j)^{-59}}{j}$

Rearranging the above,

$(1+j)^{-60}=0.00794$

$\therefore, j=0.0839$

Hence replace $j=0.0839$ in L.

$1000[\frac{1-(1.0839)^-60}{0.0839}]=11, 824.31$

The problem is that the answer is $11,820.91

I cannot grasp where I went wrong.

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    $\begingroup$ It is probably just rounding differences. $\endgroup$ – Kenny Lau Apr 17 '16 at 13:01
  • $\begingroup$ Is the calculation good,Assuming we neglect the rounding difference? @KennyLau $\endgroup$ – Tosh Apr 17 '16 at 13:10
  • $\begingroup$ You made 2 mistakes:(1) $B^p_{59}=1000\frac{1-(1+j)^{60-59}}{j}$ and not $B^p_{59}=1000\frac{1-(1+j)^{-59}}{j}$ and (2) $Pa_{\overline{60}|j}-Pa_{\overline{59}|j}=Pv^{60}$ is not the interest at time $60$ $\endgroup$ – alexjo Apr 17 '16 at 18:38
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For $0 < t < n$ the outstanding loan balance at time $t$ computed after making the $t$-th payment is $$B^p_t = P\,a_{\overline{n-t}|j}=P\,\frac{1-v^{n-t}}{j}$$ by the prospective method. So we have that the debt ad time $t$ is $D_t=B^p_t$ and the interest payed at time $t+1$ will be $I_{t+1}=jD_t=jB^p_t$, that is $$ I_{t+1}=jB^p_{t} = jP\,a_{\overline{n-t}|j}=P(1-v^{n-t}) $$

So at time $t=59$, we will have, for a loan to be repayed in $n=60$ payments of $P=1000$, $$ 7.44=I_{60} = jP\,a_{\overline{1}|j}=1000(1-v)\qquad\Longrightarrow\quad j=\frac{I_{60}}{P-I_{60}}=0.7496\% $$ The total Interest payed over the life of the loan $L$ is $$ I=nP-L=P(n-a_{\overline{n}|j}) $$ that is $$ I=1000(60-11.82091)= 11,820.91 $$

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