Question 1006280
1. Given that three numbers form an A.P. Their sum is 21,and their product is 231.find these numbers.  
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Let  x  be the middle term of this  A.P.

Then the three terms are  (x-d),  x  and  (x+d).


The sum of three is  3x = 21.  Hence,  x = 7.


The product of three is  (7-d)*7*(7+d) = 231,     or

49-d^2 = {{{231/7}}} = 33.


Hence,  d^2 = 49 - 33 = 16,  and  d = 4.


The three numbers are  7-4 = 3,  7,  and  7+4 = 11.