Question

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

• A. There exist a finite number of G.P having 27, 8 and 12 as their terms.
• B. There exist no G.P having 27, 8 and 12 as their terms.
• C. There exist an infinite number of G.P having 27, 8 and 12 as their terms.
• D. none of these

Answer 1

The correct option is C There exist an infinite number of G.P having 27, 8 and 12 as their terms.

Suppose 27, 8 and 12 are the $p$th, $q$th and $r$th terms of a G.P with first term $A$ and common ratio is $R$.
Then $27=A{R}^{p-1}\cdots \left(1\right)$
$8=A{R}^{q-1}\cdots \left(2\right)$
$12=A{R}^{r-1}\cdots \left(3\right)$
Dividing $\left(1\right)$ & $\left(3\right)$ and $\left(3\right)$ by $\left(2\right)$ then $\frac{27}{12}=R^{p-r}$
$\therefore \frac{9}{4}=R^{p-r}$
$\Rightarrow (\frac{3}{2}{)}^2=R^{p-r}\cdots (4)$
and $\frac{12}{8}=R^{r-q}$
$\therefore \frac{3}{2}=R^{r-q}\cdots \left(5\right)$
from $\left(4\right)$ and $\left(5\right)$ $(R^{r-q}{)}^2=R^{p-r}$
$\Rightarrow R^{2r-2q}=R^{p-r}$
$\Rightarrow 2r-2q=p-r$
$\Rightarrow p+2q=3r$
but $p,q,rϵ1$
$\begin{array}{ccc}p& q& r\\ 1& 1& 1\\ 2& 2& 2\\ 3& 3& 3\\ 1& 4& 3\end{array}$
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There exist an infinite number of G.P having 27, 8 and 12 as their terms.