Difference between revisions of "AoPS Wiki talk:Problem of the Day/June 16, 2011"
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==Solution== | ==Solution== | ||
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+ | The equation for the convergent sum of an infinite [[geometric sequence]] is <math>S=\dfrac{a}{1-r}</math> where ''a'' is the first term and ''r'' is the common ratio. | ||
+ | |||
+ | Therefore: | ||
+ | <cmath>2=\dfrac{k}{1-n}</cmath> and <cmath>3=\dfrac{n}{1-k}</cmath> | ||
+ | |||
+ | |||
+ | Transposing this and solving the simultaneous equations: | ||
+ | |||
+ | <cmath>2-2n=k</cmath> | ||
+ | <cmath>3-3k=n</cmath> | ||
+ | |||
+ | |||
+ | To find <math>n</math>: <cmath>3-3(2-2n)=n</cmath> | ||
+ | <cmath>\dfrac{3}{5}=n</cmath> | ||
+ | To find <math>k</math>: <cmath>2-2(\dfrac{3}{5})=k</cmath> | ||
+ | <cmath>\dfrac{4}{5}=k</cmath> | ||
+ | |||
+ | Therefore <math>k+n=\dfrac{4}{5}+\dfrac{3}{5}=\dfrac{7}{5}=1\dfrac{2}{5}</math> |
Latest revision as of 06:06, 16 June 2011
Problem
AoPSWiki:Problem of the Day/June 16, 2011
Solution
The equation for the convergent sum of an infinite geometric sequence is where a is the first term and r is the common ratio.
Therefore: and
Transposing this and solving the simultaneous equations:
To find :
To find :
Therefore