JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Probability

An increasing geometric triple is $(a,ar,a{r}^2)$ with ratio $r>1$ and all terms in $\{1,2,\dots ,40\}$. Counting by the ratio and keeping only triples whose largest term is at most $40$: for integer ratios, $r=2$ needs $4a\le 40$ ($10$ triples), $r=3$ needs $9a\le 40$ ($4$), $r=4$ needs $16a\le 40$ ($2$), $r=5$ ($1$) and $r=6$ ($1$), totalling $18$. For a ratio $r=\frac{p}{q}$ in lowest terms with $q\ge 2$ (which requires $q^2\mid a$): $r=\frac{3}{2}$ gives $(4,6,9),(8,12,18),(12,18,27),(16,24,36)$ — four; $r=\frac{5}{2}$ gives $(4,10,25)$ — one; $r=\frac{4}{3}$ gives $(9,12,16),(18,24,32)$ — two; $r=\frac{5}{3}$ gives $(9,15,25)$ — one; $r=\frac{5}{4}$ gives $(16,20,25)$ — one; $r=\frac{6}{5}$ gives $(25,30,36)$ — one, totalling $10$. The favourable count is $18+10=28$, out of $\left(\genfrac{}{}{0ex}{}{40}{3}\right)=9880$. The probability is



already in lowest terms, so $m=7$, $n=2470$ and $m+n=2477$.