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

Problem

AoPSWiki:Problem of the Day/June 16, 2011

Solution

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

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