1B. Determine the common difference of the arithmetic progression with the first term 4, in which $a_{10}, a_{31}$ and $a_{34}$ are the sides of a right triangle with hypotenuse $a_{34}$.
Solution. The members of the progression are $a_{10}=a_{1}+9 d, a_{31}=a_{1}+30 d$ and $a_{34}=a_{1}+33 d$, so if we substitute in $a_{34}^{2}=a_{10}^{2}+a_{31}^{2}$ and take into account that $a_{1}=4$, after rearrangement we obtain the quadratic equation $27 d^{2}-12 d-4=0$ whose solutions are $d_{1}=\frac{2}{3}$ and $d_{2}=-\frac{2}{9}<0$. The second solution is not possible, because for $d_{2}=-\frac{2}{9}$ we get $a_{34}=4-\frac{66}{9}=-\frac{10}{3}<0$, while the lengths of the sides of the right triangle are positive numbers.

Finally, the required value of the difference of the progression is $d=\frac{2}{3}$.