algebra precalculus – False proof that $frac{13}{6}=0$

At start, the length of a line segment is $a_0=0$. When $3$ hours have elapsed since start, its length is $a_3$. When $1$ hour has elapsed since start, its length increased by $frac{a_3}{2}$ with respect to $a_0$ (call the new length $a_1$). When $2$ hours have elapsed since start, its length increased by $frac{a_3}{3}$ (call the new length $a_2$) with respect to $a_1$. When $3$ hours have elapsed since start, its length increased by $frac{a_3}{4}$ with respect to $a_2$. What is the value of $frac{a_3}{a_1}$?

At $t=1$ hour, the length is $a_1=frac{a_3}{2}$. At $t=2$ hours, the length is $a_2=frac{a_3}{2}+frac{a_3}{3}$. At $t=3$ hours, the length is $color{red}{a_3=frac{a_3}{2}+frac{a_3}{3}+frac{a_3}{4}=frac{13}{12}a_3}$. So
$$frac{a_3}{a_1}=frac{frac{13}{12}a_3}{frac{a_3}{2}}=frac{13}{6}$$
is the answer. But notice that the red equation enables solving for $a_3$ which gives $a_3=0$. Therefore, by transitivity, $frac{a_3}{a_1}=0$.

Conclusion: $frac{13}{6}=0$

Where is the mistake?