If a,b,c are real numbers forming an A.P. and 3+a, 2+b, 3+c are in G.P. then minimum value of 'ac' is

(2+b)² = (3 +a) (3+c) 

 4+ b²+4b = 9 + ac + 3(a+c)
4+b² + 4b = 9 + 9c + 9b

ac = b² – 2b – 5
ac = (b–1)² – 6

ac ≥ -6