2. In a row, 33 weights are arranged in ascending order. It is known that any four consecutive weights can be distributed on two scales so that equilibrium is achieved. The third weight weighs 9 g, and the ninth weighs 33 g. How much does the 33rd weight weigh?
Solution. Let the weights of the weights be $a_{1}<a_{2}<\cdots<a_{33}$. For all $k=1,2, \ldots, 30$ the equalities $a_{k}+a_{k+3}=a_{k+1}+a_{k+2}$, equivalent to $a_{k+3}-a_{k+2}=$ $a_{k+1}-a_{k}$, hold. Let $a_{4}-a_{3}=a_{2}-a_{1}=d$ and $a_{5}-a_{6}=a_{3}-a_{2}=\mathrm{c}$, then $a_{6}-$ $a_{5}=a_{4}-a_{3}=d, a_{7}-a_{6}=a_{5}-a_{4}=$ c and so on. From this, $a_{1}+d+c=$ $a_{3}=9$ and $a_{1}+4 d+4 c=a_{9}=33$. From the difference of the last equalities we get $d+c=8$, and then $a_{1}=1$. We calculate $a_{33}=a_{1}+16(d+c)=129$.

Comment. An incomplete solution, brought to a system of two equations connecting $a_{1}, d, c,-4$ points.