Пожалуйста, помогите!!!

Решите задачу:

$a_1=1336, a_2=1111, \\ d=a_2-a_1=1111-1336=-225, \\ a_n=a_1+(n-1)d=1336-225(n-1)=1561-225n, \\ a_n\ \textgreater \ 0, 1561-225n\ \textgreater \ 0, \\ -225n\ \textgreater \ -1561, \\ n\ \textless \ 6 \frac{211}{225} \\ n=6, a_6=1561-225\cdot6=211.$

$a_1=-193,3,a_{105}=13, \\ a_n=a_1+(n-1)d, \\ a_{105}=-193,3+(105-1)d=-193,3+104d, \\ -193,3+104d=13, \\ 104d=206,3, \\ d=206 \frac{3}{10} :104=\frac{2063}{10}\cdot\frac{1}{104}=\frac{2063}{1040}=1\frac{1023}{1040}; \\ a_n=-193,3+(n-1)\frac{2063}{1040}=- \frac{1933}{10}-\frac{2063}{1040}+\frac{2063}{1040}n=\\=-\frac{203095}{1040}+\frac{2063}{1040}n=-\frac{40619}{208}+\frac{2063}{1040}n, \\ a_n\ \textgreater \ 0, -\frac{40619}{208}+\frac{2063}{1040}n\ \textgreater \ 0, \\ \frac{2063}{1040}n\ \textgreater \ -\frac{40619}{208},$
$n\ \textgreater \ \frac{40619}{208}\cdot \frac{1040}{2063}, \\ n\ \textgreater \ \frac{203095}{2063}, \\ n\ \textgreater \ 98 \frac{921}{2063}, \\ n=99, \\ a_{99}=-\frac{203095}{1040}+\frac{2063}{1040}\cdot99= \frac{1142}{1040} =1 \frac{51}{520}$