SAT1000 - P491

To find $a_{11}$ , just calculate it by plugging one into the function five times, as follows:

$1 \rightarrow \dfrac{1}{2} \rightarrow \dfrac{2}{3} \rightarrow \dfrac{3}{5} \rightarrow \dfrac{5}{8} \rightarrow \dfrac{8}{13}$

Thus, $a_{11}$ is $\dfrac{8}{13}$ .

Now, let's say that $a_{2010}=m$ . Then, $a_{2012}=\dfrac{1}{1+m}$ . Let's set them equal to each other and solve for m.

$a_{2010}=a_{2012}$ $m=\dfrac{1}{1+m}$ $m^{2}+m=1$ $4m^{2}+4m+1=5$ $(2m+1)^{2}=5$ $m=\dfrac{-1+\sqrt{5}}{2}$

Keep in mind that $m>0$ . Now, read this carefully. To get from $a_{2010}$ to $a_{2012}$ , we use the same process used to get from $a_{2008}$ to $a_{2010}$ . Thus, because $a_{2010}=a_{2012}$ , $a_{2008}=a_{2010}$ and $a_{2006}=a_{2008}$ and so on. All terms with even indices must be equal. Thus, $a_{20}=m$ .

Now, let's get our answer!

$a_{20}+a_{11}=\dfrac{-1+\sqrt{5}}{2}+\dfrac{8}{13} \approx 1.23341860413$

Thus, our answer is $\boxed{1233}$ .