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 $(2 q - p -1)$.
The number of ordered triplet $(p, q, r)$ is equal to

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

$p , q , r \rightarrow \text { G.P. }$
$A , AR , AR ^2 \rightarrow$ G.P. $( R \in I )$
Given, $\frac{ A -2+ AR +5+ AR ^2+4}{3}=2 AR - A -1$
$A\left(1+R+R^2\right)+7=6 A R-3 A-3 \Rightarrow A\left(R^2-5 R+4\right)=-10$
$\Rightarrow A ( R -4)( R -1)=-10 \Rightarrow A =\frac{10}{(1- R )( R -4)} \in I ^{+}$
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