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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

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answer is -6.

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Detailed Solution

(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

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