If the product of three consecutive terms of G.P. is \[216\] and the sum of the product of pair-wise is \[156\], then the numbers will be
A. \[1,3,9\]
B. \[2,6,18\]
C. \[3,9,27\]
D. \[2,4,8\]

Hint: Firstly, we suppose any three numbers and make the equation according to the question. We have two variables, that is, common ratio and first term, and we have two equations. We will get the value of the common ratio and first term by using geometric progression and putting them in the equation.

Complete step by step answer:
Firstly we will suppose the three term of geometric sequence are \[\dfrac{a}{r},a,ar\] where the middle term is \[a\] and \[r\] be the common ratio
Now, we will get the equation according to the question
\[\begin{array}{l}\dfrac{a}{r} \times a \times ar = 216\\{a^3} = 216\\{a^3} = {(6)^3}\\a = 6\end{array}\]
Here, we will get the second equation
\[\begin{array}{l}\left( {\dfrac{a}{r} \times a} \right) + \left( {a \times ar} \right) + \left( {\dfrac{a}{r} \times ar} \right) = 156\\\dfrac{{{a^2}}}{r} + {a^2}r + {a^2} = 156\\{a^2}\left( {\dfrac{{1 + {r^2} + r}}{r}} \right) = 156\\{r^2} + r + 1 = \dfrac{{156 \times r}}{{{{(6)}^2}}}\end{array}\]
Now, we will solve further to get the quadratic equation
\[\begin{array}{l}{r^2} + r + 1 = \dfrac{{13r}}{3}\\3{r^2} + 3r + 3 = 13r\\3{r^2} + 3r - 13r + 3 = 0\\3{r^2} - 10r + 3 = 0\end{array}\]
Here, we will factorize the obtained quadratic equation
\[\begin{array}{l}r = \dfrac{{10 \pm \sqrt {100 - 36} }}{{2(3)}}\\r = \dfrac{{10 \pm \sqrt {64} }}{6}\\r = \dfrac{{10 + 8}}{6},\dfrac{{10 - 8}}{6}\\r = 3,\dfrac{1}{3}\end{array}\]
Now we will get value of three consecutive terms by putting the value of \[a\] and \[r = 3\]
\[\begin{array}{l}\dfrac{a}{r},a,ar = \dfrac{6}{3},6,6 \times 3\\\dfrac{a}{r},a,ar = 2,6,18\end{array}\]
Now we will get value of three consecutive terms by putting the value of \[a\] and \[r = \dfrac{1}{3}\]
\[\begin{array}{l}\dfrac{a}{r},a,ar = \dfrac{6}{{\dfrac{1}{3}}},6,6 \times \dfrac{1}{3}\\\dfrac{a}{r},a,ar = 18,6,2\end{array}\]
Hence, the value of the three consecutive terms \[\dfrac{a}{r},a,ar\] is \[18,6,2\]
Therefore, the option (B) is correct.

Note: We are given information about three consecutive terms of a geometric progression and we would have normally taken three consecutive terms as \[a,ar,a{r^2}\] but to solve the given question we need to eliminate \[r\] , that’s why the form \[\dfrac{a}{r},a,ar\] is used as it eliminates \[r\] and directly gives the value of \[a\] and thus the question is easily solved.