# A sequence {a_n} is defined recursively, with a_1 = 1, a_2=2 and, for n>2, a_n = (a_(n-1))/(a_(n-2)). How do you find the term a_(241)?

Jun 7, 2016

${a}_{241} = 1$

#### Explanation:

First let's observe some values of the sequence to find a common trend.

${a}_{3} = \frac{{a}_{2}}{{a}_{1}} = \frac{2}{1} = 2$
${a}_{4} = \frac{{a}_{3}}{{a}_{2}} = \frac{2}{2} = 1$
${a}_{5} = \frac{{a}_{4}}{{a}_{3}} = \frac{1}{2}$
${a}_{6} = \frac{{a}_{5}}{{a}_{4}} = \frac{\frac{1}{2}}{1} = \frac{1}{2}$
${a}_{7} = \frac{{a}_{6}}{{a}_{5}} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1 = {a}_{1}$
${a}_{8} = \frac{{a}_{7}}{{a}_{6}} = \frac{1}{\frac{1}{2}} = 2 = {a}_{2}$

As you probably notice now, the sequence repeats every $6$ terms, meaning that ${a}_{1} = {a}_{7} = {a}_{13} = \ldots {a}_{1 + 6 t}$.

In fact, we can generalize. If $1 \le k < 7$, then

${a}_{k} = {a}_{k + 6} = {a}_{k + 12} = \ldots = {a}_{k + 6 t}$.

What this means that for any $n$, we can find which value ${a}_{n}$ corresponds to by taking the remainder when we divide $n$ by $6$.

In the case of $n = 241$, $\frac{241}{6} = \frac{240 + 1}{6} = 40 \frac{1}{6}$, which implies a remainder of $1$ and thus ${a}_{241}$ corresponds to ${a}_{1}$, which is $1$.