Question 711100
Let's call the three integers in the original series a, b, and c.  Because this is a geometric series, we know that
 
{{{b/a = c/b}}}, so multiplying both sides by {{{a*b}}} gives
{{{b^2 = ac}}}
{{{c = b^2/a}}}
 
Adding 12 to the second number (b) gives an arithmetic series, so we know that
 
{{{b+12 - a = c - (b+12)}}}, or
{{{a + c = 2b+24}}}
 
And finally, adding 96 to the third term of the new series gives us another geometric series, so we have
 
{{{(b+12)/a = (c+96)/(b+12)}}}, so multiplying both sides by {{{a*(b+12)}}} gives
 
{{{(b+12)^2 = a*(c+96)}}}
{{{b^2 + 24b + 144 = ac + 96a}}}
 
and since {{{b^2 = ac}}}, we have
 
{{{24b + 144 = 96a}}}
{{{b = 4a-6}}}
 
And using {{{c = b^2/a}}}
 
{{{c = (4a-6)^2/a}}}
{{{c = (16a^2-48a+36)/a}}}
{{{c = 16a - 48 + 36/a}}}
 
From the equation above, c is only an integer when a is a factor if 36, so we can check the nine possibilities:
 
(1, -2, 4)
(2, 2, 2)
(3, 6, 12)
(4, 10, 25)
(6, 18, 54)
(9, 30, 100)
(12, 42, 147)
(18, 66, 242)
(36, 138, 529)
 
Of these 9 geometric series, only one generates an arithmetic series when 12 is added to the second term:  (6, 18, 54)
 
Adding 12 to the second term gives the arithmetic series (6, 30, 54), 
 
And adding 96 to the third term gives the geometric series (6, 30, 150).