[tex]S_{12} = 120\\a_1 = 4\\n = 12\\r = ?\\a_{12} = ?\\\\S_{n} = \frac{2a_1+(n-1)r}{2}\cdot n\\\\120 = \frac{2\cdot4 + (12-1)\cdot r}{2} \cdot12\\\\120 = (8+11r)\cdot6 \ \ /:6\\\\20 = 8+11r\\\\20-8 = 11r\\\\12 = 11r \ \ /:11\\\\r = \frac{12}{11}\\\\\boxed{r = 1\frac{1}{11}}[/tex]
[tex]a_{n} = a_1 + (n-1)\cdot r\\\\a_{12} = 4+(12-1)\cdot\frac{12}{11}\\\\a_{12} = 4+11\cdot\frac{12}{11}\\\\a_{12} = 4+12}\\\\\boxed{a_{12} = 16}[/tex]
