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A loan of \$10 000 repaid over a period of $n$ years at $r$% per year satisfies the equation $$ 10000 = \frac{x}{\left(1+\frac{r}{100}\right)}+\frac{x}{\left(1+\frac{r}{100}\right)^2}+\frac{x}{\left(1+\frac{r}{100}\right)^3}+\cdots++\frac{x}{\left(1+\frac{r}{100}\right)^n} $$ where $x$ is the repayment installment. Find $x$ in terms of $r$ and $n$ and compute its value if $r$ = 10 and $n$ = 20.

I've been doing loads of series exercises but this one now has more than 1 variable and now I feel stumped. How do you get started with this?

This is not a p-series because the numerator is not 1? Based on the next question this might be a harmonic series...

Any assistance would be great.

Thanks

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    $\begingroup$ Write $z=\frac{1}{1+\frac{r}{100}}$. $\endgroup$
    – Thomas Andrews
    Commented Nov 20, 2012 at 13:39
  • $\begingroup$ You can take the variable $x$ out since it occurs in every term. So it's really only a sum (not a series, since it's only got finitely many terms) in one variable. $\endgroup$
    – John Gowers
    Commented Jun 20, 2013 at 21:27

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$$\frac{10000 (1+\frac r{100})^n}x=1+(1+\frac r{100})+(1+\frac r{100})^2+\cdots+(1+\frac r{100})^{n-1}=\frac{(1+\frac r{100})^n-1}{(1+\frac r{100})-1}$$

or, $$\frac{10000\cdot \frac r{100}}x=\frac{(1+\frac r{100})^n-1}{(1+\frac r{100})^n}$$

or, $$x=\frac{100r(1+\frac r{100})^n}{(1+\frac r{100})^n-1}$$

$$=\frac{1000 (\frac{11}{10})^{20}}{(\frac{11}{10})^{20}-1}= \frac{1000}{1-(\frac{10}{11})^{20}}$$ as $r=10,n=20$

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