Let the four terms be 12−3k, 12−k, 12+k, 12+3k with common difference l=2k.
Sum =48 is satisfied. Product condition:
(144−9k2)(144−k2)+16k4=361
25k4−1440k2+20375=0
Let u=k2: 25u2−1440u+20375=0
u=501440±2073600−2037500=501440±190
u=25 (integer) or u=32.6 (rejected).
k=5, l=10. Terms: −3,7,17,27.
Largest term =27.