Question

Does there exist a geometric progression containing $27,8$ and $12$ as three of its terms ? If it exists, how many such progressions are possible?

Answer 1

Let is possible $8$ be the first term and $12$ and $27$ be $mth$ and $nth$ terms repectively.
$\therefore 12={ar}^{m-1}={8r}^{m-1},27={8r}^{n-1}$
$\therefore \frac{3}{2}=r^{m-1},{\left(\frac{3}{2}\right)}^3=r^{n-1}=r^{3(m-1)}$
$\therefore n-1=3m-3$ or $3m=n+2$
or $\frac{m}{1}=\frac{m+2}{3}=k$ say $\therefore m=k,n=3k-2$.
By giving $k$ different values we get integral values of $m$ and $n$. Hence there can be infinite number of $G.P.s$ whose any three terms will be $8,12,27$ (not consecutive).