Question

Let $p,q,r$ be positive integers such that $\frac{r}{q}$ is an integer. Also $p,q,r$ (taken in that order) are in geometric progression and the arithmetic mean of $p-2,q+5,r+4$ is $(2q-p-1)$.
The number of ordered triplet $(p,q,r)$ is equal to

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

$p,q,r\to \text{ G.P. }$
$A,AR,A{R}^2\to$ G.P. $(R\in I)$
Given, $\frac{A-2+AR+5+A{R}^2+4}{3}=2AR-A-1$
$A\left(1+R+R^2\right)+7=6AR-3A-3\Rightarrow A\left(R^2-5R+4\right)=-10$
$\Rightarrow A(R-4)(R-1)=-10\Rightarrow A=\frac{10}{(1-R)(R-4)}\in I^{+}$
Since $A$ is a positive integer
$\therefore R=2\text{ or }R=3$
If $R=2$ and $R=3$, then $A=5$
$\therefore$ Value of $p=5$
Numbers are $5,10,20$ or $5,15,45$
$\therefore$ Total number of ordered triplets of $(p,q,r)$ are 2