Notes: given a loan on $6000, interest is 2%, how much would have to be paid monthly in order to pay off the loan? i know the question is really simple, but...

Saturday, 17 January 2015

given a loan on $6000, interest is 2%, how much would have to be paid monthly in order to pay off the loan? i know the question is really simple, but...

The principal
(P) is $6000 and the interest rate (r) is 0.02.

Now if the loan was for 1
month we would have

$\displaystyle{L}={P}{\left({1}+{r}\right)}$ or $\displaystyle{P}=\frac{{L}}{{{1}+{r}}}$

If the loan
was for 2 months we would have

$\displaystyle{\left({1}+{r}\right)}{\left({\left({1}+{r}\right)}{P}-{L}\right)}={L}$ or $\displaystyle{{\left({1}+{r}\right)}}^{{2}}{P}-{\left({1}+{r}\right)}{L}={L}$
which gives $\displaystyle{P}=\frac{{L}}{{{1}+{r}}}+\frac{{L}}{{{\left({1}+{r}\right)}}^{{2}}}$

So if the loan is n months we would
have

$\displaystyle{P}={\sum_{{{j}={1}}}^{{n}}}\frac{{L}}{{{\left({1}+{r}\right)}}^{{j}}}={L}{\sum_{{{j}={1}}}^{{n}}}\frac{{1}}{{{\left({1}+{r}\right)}}^{{n}}}$

We can solve this by noting that

$\displaystyle{\left({1}+{r}\right)}{P}={L}{\sum_{{{j}={0}}}^{{{n}-{1}}}}$
1/(1+r)^n

$\displaystyle{\left({1}+{r}\right)}{P}-{P}={r}{P}={L}{\left({1}-\frac{{1}}{{{\left({1}+{r}\right)}}^{{n}}}\right)}$

So $\displaystyle{L}=\frac{{{r}{P}}}{{{1}-\frac{{1}}{{{\left({1}+{r}\right)}}^{{n}}}}}$

Substituting P = $6000, r = 0.02 and n =
24

$\displaystyle{L}=\frac{{{0.02}{\left({6000}\right)}}}{{{1}-\frac{{1}}{{{\left({1}-{0.02}\right)}}^{{24}}}}}$

Evaluating we
get $317.23 per month. Now this does seem high since 6000/24=250 but 2% per month is 48% for
the entire loan, so this is actually a high interest rate made more attractive by stating it as
a monthly interest rate.

So the final answer is $317.23 will pay off the loan
in 24 months.

Notes

Saturday, 17 January 2015

given a loan on $6000, interest is 2%, how much would have to be paid monthly in order to pay off the loan? i know the question is really simple, but...

$\displaystyle{L}=\frac{{{0.02}{\left({6000}\right)}}}{{{1}-\frac{{1}}{{{\left({1}-{0.02}\right)}}^{{24}}}}}$

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