[Algebra]
Problem: A list $a_{1}, a_{2}, a_{3}, a_{4}$ of rational numbers is defined so that if one term is equal to $r$, then the next term is equal to $1+\frac{1}{1+r}$. For example, if $a_{3}=\frac{41}{29}$, then $a_{4}=1+\frac{1}{1+(41 / 29)}=\frac{99}{70}$. If $a_{3}=\frac{41}{29}$, what is the value of $a_{1} ?$

Solution: If $r$ is a term in the sequence and $s$ is the next term, then $s=1+\frac{1}{1+r}$.

This means that $s-1=\frac{1}{1+r}$ and so $\frac{1}{s-1}=1+r$ which gives $r=\frac{1}{s-1}-1$.

Therefore, since $a_{3}=\frac{41}{29}$, then

$$
a_{2}=\frac{1}{a_{3}-1}-1=\frac{1}{(41 / 29)-1}-1=\frac{1}{12 / 29}-1=\frac{29}{12}-1=\frac{17}{12}
$$

Further, since $a_{2}=\frac{17}{12}$, then

$$
a_{1}=\frac{1}{a_{2}-1}-1=\frac{1}{(17 / 12)-1}-1=\frac{1}{5 / 12}-1=\frac{12}{5}-1=\frac{7}{5}
$$

Final answer: $\frac{7}{5}$