Odpowiedź:
[tex]a) a_{4} = a_{1} + (4- 1)r \\ 6 = a_{1} + 3r \\ \\ a_{8} = a_{1} + (8- 1)r \\ 14 = a_{1} + 7r \\ \\ 6 = a_{1} + 3r \\ 14 = a_{1} + 7r \\ \\ 6 - 3r = a_{1} \\ 14 = 6 - 3r + 7r \\ \\ 6 - 3r = a_{1} \\ 8= 4r \\ \\6 - 3r = a_{1} \\ r = 2 \\ \\ 6 - 3 \times 2 = a_{1} \\ r = 2 \\ \\ 6 - 6 = a_{1} \\ r = 2 \\ \\ a_{1} = 0 \\ r = 2[/tex]
