If the sum of three numbers in G.P is 21 and their product is 216, then the numbers are 

Correct option is (B) 3, 6, 12

Let \(\frac ar,a,ar\) be three required numbers that are in G.P.

Given that their product is 216.

\(\therefore\) \(\frac ar.a.ar=216\)

\(\Rightarrow a^3=216=6^3\)

\(\Rightarrow\) a = 6

Also given that their sum is 21.

\(\therefore\) \(\frac ar+a+ar=21\)

\(\Rightarrow\frac6r+6+6r=21\)

\(\Rightarrow6r^2+6r+6=21r\)

\(\Rightarrow6r^2-15r+6=0\)

\(\Rightarrow2r^2-5r+2=0\)

\(\Rightarrow2r^2-4r-r+2=0\)

\(\Rightarrow2r(r-2)-1(r-2)=0\)

\(\Rightarrow(r-2)(2r-1)=0\)

\(\Rightarrow r-2=0\;or\;2r-1=0\)

\(\Rightarrow r=2\;or\;r=\frac12\)

\(\therefore\frac ar=\frac62=3\;or\;\frac ar=\frac6{\frac12}=6\times2=12\)

\(ar=6.2=12\;or\;ar=6\times\frac12=3\)

Hence, required numbers are 3, 6, 12.

Correct option is B) 3, 6, 12