Question 1190791
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a1 + a2 + a3 = 69

Since we are given a1,a2,a3 form an AP, we can write:

a1 + (a1+k) + (a1+2k) = 69    where k is the common difference

3a1 + 3k = 69
a1 + k = 23   

Trying  a1 = 21,  a1+k = 23, and a1+2k = 25,  we see this AP meets the requiremnt that the first two terms multiply out to 483,  thus  {21,23,25} satisfies the condition.

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Alternate solution:
a1 * (a1+k) = 483
{{{a1^2 + a1*k - 483 = 0}}}
{{{ (a1-21)(a1+23) = 0 }}}
Expanding it back to a quadratic:
{{{ a1^2 +2a1 - 483 = 0 }}}  so we see  a1=21, k=2 works, resulting in {21,23,25} as above