Question

Задание во вложении.

Answer 1

$a_{3} = 7\\ a_{19} = 55 \\ \\ a_{n} = a_{1} + (n - 1)d \\ \\ a_{3} = a_{1} + (3 - 1)d \\ a_{19} = a_{1} + (19 - 1)d \\ \\ 7 = a_{1} + 2d \: \: \: | \times ( - 1) \\ 55 = a_{1} + 18d \\ \\ - 7 = - a_{1} - 2d \\ 55 = a_{1} + 18d \\ \\ 55 - 7 = 18d - 2d \\ 16d = 48 \\ d = 48 \div 16 \\ d = 3$

Answer 2

$\displaystyle\bf\\\left \{ {{a_{19} =55} \atop {a_{3} =7}} \right. \\\\\\-\left \{ {{a_{1}+18d=55 } \atop {a_{1} +2d=7}} \right. \\----------\\16d=48\\\\d=48:16\\\\\boxed{d=3}$