Question 1023746
Let a be the first term and d be the common difference.


*[tex \large a + (a+d) + \ldots + (a+4d) + 90 = (a+5d) + (a+6d) + \ldots (a+9d)]


The a's cancel, leaving *[tex \large d + 2d + 3d + 4d + 90 = 5d + 6d + 7d + 8d + 9d] or *[tex \large 10d + 90 = 35d]. Solving, *[tex \large d = \frac{90}{25} = \frac{18}{5}].