[tex]Zadanie\\\\a_1=\frac{1}{4}\\\\a_2=\frac{3}{4}\\\\r=a_2-a_1\\\\r=\frac{3}{4}-\frac{1}{4}= \frac{2}{4}=\frac{1}{2}\\\\a_n=a_1+(n-1)r\\\\a_9=\frac{1}{4}+(9-1)\cdot\frac{1}{2}= \frac{1}{4}+8\cdot\frac{1}{2}=\frac{1}{4}+4=4\frac{1}{4}[/tex]
