A sequence of numbers have the property that x, 12, y, where x > 0, y > 0, form a geometric sequence while 12, x, 3y form an arithmetic sequence. A)If xy = k, find k. B)Find the value of x and y.

A) Since x, 12, y form a geometric sequence, we know that there exists some real number "r" (common ratio) such that 12=xr and y=12r. Hence, r= 12/x =y/12 => xy=144 => k=144 (cross multiply).B)We already have one equation involving both x and y (xy=144) so since there are two unknown variables we aim to obtain another one. Using the fact that 12, x, 3y form an arithmetric sequece, we know that there exists a real number "d" (common difference) such that x=12+d and 3y=x+d, therefore d=x-12=3y-x => 12+3y=2x. Using that xy=144, we substitute x=144/y into the latter equation to get 12+3y=288/y =>3y^2 +12y-288=0, upon multiplying both sides by y. Finally, we divide both sides by 3 to obtain the quadratic: y^2+4y-96=0, which has discriminant 4^2-4*(1)*(-96)=400 and therefore y=-2+-10 => y=-12 or y=8. The condition y>0 allows us to deduce that y=8. Using xy=144, we calculate x=144/8=18.