Mathematics - Probability Question with Solution | TestHub

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