Three positive numbers from a geometric sequence .If the geometric mean of the first two numbers is 6 and the geometric mean of the last two numbers is 24, find the three numbers and their common ratio.
Answer (1)
SOLUTION:We leta₁ = first numbera₂ = second numbera₃ = third numberBased on the problem, the geometric sequence formed isa₁, 6, a₂, 24, a₃Solving for a₂ by using the concept of the common ratio of geometric sequence[tex]\begin{aligned} \frac{a_2}{6} & = \frac{24}{a_2} \\ a_2^2 & = 6(24) \\ a_2^2 & = 144 \\ \sqrt{a_2^2} & = \sqrt{144} \\ a_2 & = \boxed{12} \end{aligned}[/tex]Calculating a₁ and a₃• For a₁[tex]\begin{aligned} \frac{6}{a_1} & = \frac{a_2}{6} \\ \frac{6}{a_1} & = \frac{12}{6} \\ \frac{6}{a_1} & = 2 \\ a_1 & = \frac{6}{2} \\ a_1 & = \boxed{3} \end{aligned}[/tex]• For a₃[tex]\begin{aligned} \frac{a_3}{24} & = \frac{24}{a_2} \\ \frac{a_3}{24} & = \frac{24}{12} \\ \frac{a_3}{24} & = 2 \\ a_3 & = (24)(2) \\ a_3 & = \boxed{48} \end{aligned}[/tex]Computing for the common ratio[tex]\begin{aligned} r & = \frac{6}{a_1} \\ & = \frac{6}{3} \\ & = \boxed{2} \end{aligned}[/tex]Hence, the three positive integers and the common ratio are 3, 12, 48, and 2, respectively.
