AoPS Wiki talk:Problem of the Day/June 16, 2011

Problem

The equation for the convergent sum of an infinite geometric sequence is $S=\dfrac{a}{1-r}$ where a is the first term and r is the common ratio.

Therefore: $2=\dfrac{k}{1-n}$ and $3=\dfrac{n}{1-k}$

Transposing this and solving the simultaneous equations:

$2-2n=k$ $3-3k=n$

To find $n$ : $3-3(2-2n)=n$ $\dfrac{3}{5}=n$ To find $k$ : $2-2(\dfrac{3}{5})=k$ $\dfrac{4}{5}=k$

Therefore $k+n=\dfrac{4}{5}+\dfrac{3}{5}=\dfrac{7}{5}=1\dfrac{2}{5}$