Question: Does there exists a GP containing 8, 12 and 27 as three of its terms? If it exists how many such progressions are possible?
What I have tried: First of all I have made three equations as follows Here 27, 12, 8 are $p^{th},q^{th},r^{th} $ term of GP
$27=AR^{p-1} ...(1)$
$12=AR^{q-1} ....(2)$
$8=AR^{r-1} ....(3)$
Then I divided equation 1 by 2 and equation 2 by 3 and got the below equations.
$\frac{27}{12}=(\frac{3}{2})^2=R^{p-q} ....(4)$
$\frac{12}{8}=\frac{3}{2}=R^{q-r} ....(5)$
Then I got two more equations as follows
$p-q=2$
$q-r=1$
Then I have let $p=k$ and got $q=k-2$ and $r=k-3$
Now for each value of k we get different values of $q$ and $r$.
Now my question is that, is this enough to prove there can be innumerable progressions containing 27,8,12 as its terms?
There are infinitely many such progressions. To see this, note that $12 = 1.5 \times 8$, and $27 = 1.5 \times 1.5 \times 8$.
Now have $a_n=8\cdot(1.5^{\frac{1}{q}})^n$ for some $q \in \mathbb N$. It will have 8 as the 0th therm, 12 as the $q$th term and 27 as the $3q$th term.
Your answer is also correct. It could be written as $a_n=(8\cdot 1.5^p) \times 1.5^n$ for some $p \in \mathbb N$.
Together with my answer you can even find all such progressions, they have the form $a_n=(8\cdot 1.5^{\frac{p}{q}}) \times (1.5^{\frac{1}{q}})^n$, for some $p,q \in \mathbb N$.