Question 383150
let z1 = a
z2 = ar, and
{{{z3 = ar^2}}}.  Then from the given,
{{{(a+ar+ar^2)/3 = 10}}}, and {{{(a^2+a^2r^2 + a^2r^4)/3 = 20i}}}, both arising from the given.
Hence
{{{a+ar+ar^2 = 30}}},            (1)   and 
{{{a^2(1+r^2 + r^4) = 60i}}}.    (2)
Divide (2) by (1), to get
{{{(a^2(1 + r^2 + r^4))/(a(1+r+r^2)) = (60i)/(30)}}}, or
{{{a(1-r + r^2) = 2i}}}, or {{{a-ar + ar^2 = 2i}}}.  Subtract this result from
{{{a+ar+ar^2 = 30}}}, to get
{{{2ar = 30 - 2i}}}, OR, {{{ar = 15 - i}}}.  BUT ar is exactly the middle term of the geometric sequence of 3 geometric terms, or z2.  
Thus, z2 = 15 - i.