How many geometric progressions is / are possible containing 27, 8 and 12 as three of its / their terms?
${\text{A}}{\text{. }}$ One
${\text{B}}{\text{. }}$ Two
${\text{C}}{\text{. }}$ Four
${\text{D}}{\text{.}}$ Infinitely many

Hint: Here nothing is mentioned about which term is given so we can take any term as any term and then we have to apply the rules of GP so that we can find terms like common ratio and which term is what.

Complete step-by-step answer:
Here we have not given that which term of GP is what.
So let’s say
First term = $a= 8 $ …………………………(eq.1)
And we know that ${{\text{P}}^{th}}$ of GP is $a{r^{p - 1}}$
So assume ${{\text{P}}^{th}}$ term is = 12 = $a{r^{p - 1}}$ …………….(eq.2)
And as like ${{\text{Q}}^{th}}$ term is = 27 = $a{r^{q - 1}}$ ………………(eq.3)
On dividing eq.2 by eq.1 we get,
$\dfrac{{a{r^{p - 1}}}}{a} = \dfrac{{12}}{8}$
$\therefore {r^{p - 1}} = 1.5$ ………………………..( eq.4)
Now on dividing eq.3 by eq.1 we get,
$\dfrac{{a{r^{q - 1}}}}{a} = \dfrac{{27}}{8}$
${r^{q - 1}} = 3.375 = {\left( {1.5} \right)^3}$ …………………..(eq.5)
Now on dividing eq.5 by eq.4 we get,
$\dfrac{{{r^{q - 1}}}}{{{r^{p - 1}}}} = \dfrac{{{{\left( {1.5} \right)}^3}}}{{\left( {1.5} \right)}} = {\left( {1.5} \right)^2}$
${r^{q - p}} = {\left( {1.5} \right)^2}$
Putting the value from eq.4 we get,
${r^{q - p}} = {r^{2p - 2}}$
On further calculation we get,
$Q = 3p -2 $
This is the only condition which should be satisfied. Hence we can make infinite GP from 8 , 12 and 27.
Hence option D is the correct option.

Note: Whenever we get this type of question the key concept of solving is we have to assume the terms and find the relation between them so that we can make any idea about how many patterns of this type can be made.