Three distinct numbers are selected randomly from the set $\{1,2,3, \ldots \ldots, 40\}$. If the probability, that the selected numbers are in an increasing G.P. is $\frac{m}{n}$, $\operatorname{gcd}(m, n)=1$, then $m+n$ is equal to _____.

Question

Aicharya · Accepted Answer

$1 \leq \mathrm{a} \lt \mathrm{ar} \lt \mathrm{ar}^2 \leq 40$ (If $r \in N$ ) If $r=2$ $1 \leq a \lt 2 a \lt 4 a \leq 40$ $a \in\{1, \ldots \ldots ., 10\}$ ________ (10 GP) If $r=3$ $1 \leq a \lt 3 a \lt 9 a \leq 40$ $\mathrm{a} \in\{1,2,3,4\}$ ________ ________ (4 GP) If $\mathrm{r}=4$ $1 \leq a \lt 4 a \lt 16 a \leq 40$ $a \in\{1,2\}$ ________ ________ (2 GP) If $r=5$ $1 \leq a \lt 5 a \lt 25 a \leq 40$ $a \in\{1\}$ ________ ________ (1 GP) If $r=6$ $1 \leq a \lt 6 a \lt 36 a \leq 40$ $\mathrm{a} \in\{1\}$ ________ ________ (1 GP) $\left(\mathrm{P}=\frac{18}{9880}=\frac{9}{4940}ight)$ as per NTA for $\mathrm{r} \in \mathrm{N}$ $\mathrm{m}+\mathrm{n}=4949$ If $\underline{r} 
otin \mathrm{~N}$ (also possible) $\mathrm{r}=\frac{3}{2}$ $\operatorname{ar}^2=\frac{9 \mathrm{a}}{4} ; \mathrm{a}=4 \mathrm{k}$ $\left.\begin{array}{l}(4,6,9) \ (8,12,18) \ (12,18,27) \ (16,24,36)\end{array}ight\} 4 \mathrm{GP}$ $\begin{aligned} & \begin{array}{l}\mathrm{r}=\frac{5}{2} \quad \mathrm{ar}^2=\frac{25 \mathrm{a}}{4} ; \mathrm{a}=4 \mathrm{k} \ \ \ (4,10,25) \ldots \ldots . .(1) \mathrm{GP} \ \mathrm{r}=\frac{4}{3} \quad \mathrm{ar}^2=\frac{16 \mathrm{a}}{9} ightarrow \mathrm{a}=9 \mathrm{k} \ \ \ (9,12,16),(18,24,32) \ldots \ldots .(2) \mathrm{GP} \ \mathrm{r}=\frac{5}{3} \quad \mathrm{ar}^2=\frac{25 \mathrm{a}}{9} ; \mathrm{a}=9 \mathrm{k} \ \ \quad(9,15,25) \ldots \ldots \ldots .(1) \mathrm{GP} \ \mathrm{r}=\frac{5}{4} \quad \mathrm{ar}^2=\frac{25 \mathrm{a}}{16} ; \mathrm{a}=16 \mathrm{k}\end{array} \ & \quad \begin{array}{l}(16,20,25) \ldots \ldots \ldots . .(1) \mathrm{GP} \ \mathrm{r}=\frac{6}{5} \quad \mathrm{ar}^2=\frac{36 \mathrm{a}}{25} ; \mathrm{a}=25 \mathrm{k}\end{array} \ & \begin{array}{l}(25,30,36) \ldots \ldots \ldots . .(1) \mathrm{GP} \ \mathrm{Total}=18+10=28 \ \mathrm{P}=\frac{28}{40} \mathrm{C}_3=\frac{28}{9880}=\frac{7}{2470} \ \mathrm{~m}+\mathrm{n}=2477\end{array}\end{aligned}$

Three distinct numbers are selected randomly from the set $\{1,2,3, \ldots \ldots, 40\}$. If the probability…

Solution