Discussion
The sum of three numbers in an A.P. is 27 and their product is 504. Find the numbers.
Answer: A
Let the three numbers be \(a-d, a, a+d\). Their sum is \((a-d)+a+(a+d) = 3a = 27\), so \(a=9\). Their product is \((a-d)a(a+d) = a(a^2-d^2) = 504\). Substituting a=9, we get \(9(81-d^2)=504\). \(81-d^2 = 504/9 = 56\). \(d^2 = 81-56 = 25\), so \(d = \pm 5\). If d=5, the numbers are 4, 9, 14. If d=-5, the numbers are 14, 9, 4. The numbers are 4, 9, 14.
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