smc

We can separate this into two cases. Case 1: $a_1=0$ Here, clearly the minimum sum is when all of the $a_i$ are $0$, so $0\le x$. The maximum sum is when the rest of the $a_i$ are all $2$. In this case, $$\sum _{n=2}^{25}\frac{2}{3^n}<2\sum _{n=2}^{\infty }\frac{1}{3^n}=2\left(\frac{\frac{1}{9}}{1-\frac{1}{3}}\right)=\frac{1}{3}.$$ Therefore, in this case, $0\le x<1/3$. Case 2: $a_1=2$ Here, again, the minimum sum is when the rest of the $a_i$ are all $0$, in which case the minimum sum is $\frac{2}{3}$. The maximum sum is attained when all of the $a_i$ are $2$. In that case, $$\sum _{n=1}^{25}\frac{2}{3^n}<2\sum _{n=1}^{\infty }\frac{1}{3^n}=2\left(\frac{\frac{1}{3}}{1-\frac{1}{3}}\right)=1.$$ Therefore, in this case, $2/3\le x<1$. Thus, we must have $0\le x<1/3$ or $2/3\le x<1$, $\boxed{0\le x<1/3\text{ or }2/3\le x<1}$.