Ответ: a₁=11,   n=5.

Объяснение:

[tex]\displaystyle\\d=3\ \ \ \ a_n=23\ \ \ \ S_n=85\ \ \ \ \ \ n\in\mathbb N\\\\\left \{ {{a_n=a_1+(n-1)*d=23} \atop {S_n=\frac{a_1+a_n}{2}*n=85 }} \right.\ \ \ \ \left \{ {a_1+nd-d=23} \atop {\frac{a_1+a_n}{2}*n=85\ |*2 }} \right. \ \ \ \ \left \{ {a_1+3n-3=23} \atop {(a_1+a_n)*n=170}} \right. \\\\\\\left \{ {{a_1=26-3n} \atop {(26-3n+23)*n=170}} \right. \ \ \ \ \left \{ {{a_1=26-3n} \atop {(49-3n)*n=170}} \right. \ \ \ \ \left \{ {{a_1=26-3n} \atop {49n-3n^2=170}} \right.\\\\\[/tex]

[tex]\displaystyle\\\left \{ {{a_1=26-3n} \atop {3n^2-49n+170=0}} \right.\ \ \ \ \left \{ {{a_1=26-3n} \atop {3n^2-15n-34n+170=0}} \right.\ \ \ \ \\\\\\\left \{ {{a_1=26-3n} \atop 3n*(n-5)-34*(x-5)=0}} \right. \ \ \ \ \left \{ {{a_1=26-3n} \atop {(n-5)*(3n-34)=0}} \right.\ \ \ \\\\\\\ \left \{ {{a_1=11\ \ } \atop {n_1=5\in\ \ n_2=\frac{34}{3} \notin}} \right. .\ \ \ \ \ \ \Rightarrow\\\\a_1=11\ \ \ \ \ \ n=5.[/tex]