Question 388620
{{{U_(n+1) = (a/4)U_n + 12}}}
a. U_0 = 16 ===> {{{U_1 = (a/4)*16 + 12 = 4a + 12}}}.  Then
{{{U_2 = (a/4)*(4a+12) + 12 = a(a+3)+12 = a^2 + 3a + 12}}}.

b.If {{{U_2 = 30 = a^2 + 3a + 12}}}, then {{{a^2 + 3a -18 = 0}}}, and (a+6)(a-3) = 0.  Therefore a = 3.  (Eliminate -6 because we want a>0).