Three consecutive in G.P.: (\dfrac{a}{r},\ a,\ ar)

Sum: (a\left(\dfrac{1}{r}+1+r\right)=26)

Product: (a^3=216\implies a=6)

[ 6\left(\dfrac{1}{r}+1+r\right)=26\implies\dfrac{1}{r}+r=\dfrac{13}{3}-1=\dfrac{10}{3} ] [ r+\dfrac{1}{r}=\dfrac{10}{3}\implies3r^2-10r+3=0\implies(3r-1)(r-3)=0 ] [ r=3\text{ or }\dfrac{1}{3} ]

Numbers: (2,6,18)

Answer: (2,6,18)