Question

11 и 12 задание .Помогите пожалуйста

Answer 1

Решение на фото........

Answer 2

$11) a_1 = -8; a_2 = -1; a_3 = 6...\\d = a_2 - a_1 = -1 + 8 = 7\\a_n = a_1 + d(n-1)\\a_{51} = a_1 + d(50)\\a_{51} = 342\\12) (\frac1{4a}+ \frac1{8a})*\frac{a^2}2 = \frac{3a^2}{16a} = -\frac{3*1.6}{16} = -0.3$