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.
