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(D) The total number of ways to select $3$ distinct numbers from $40$ is $^{40}C_3 = \frac{40 	imes 39 	imes 38}{3 	imes 2 	imes 1} = 9880$.Let th

Vedclass · Accepted Answer

$2477$

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(D) The total number of ways to select $3$ distinct numbers from $40$ is $^{40}C_3 = \frac{40 	imes 39 	imes 38}{3 	imes 2 	imes 1} = 9880$.Let the numbers be $a, ar, ar^2$ where $r = \frac{p}{q}$ in simplest form $(p > q \geq 1)$.Then the numbers are $a, a(\frac{p}{q}), a(\frac{p^2}{q^2})$,which implies $a$ must be a multiple of $q^2$. Let $a = k q^2$.The numbers are $k q^2, k q p, k p^2$.Since $k p^2 \leq 40$,we test values for $p$ and $q$ $(p > q)$:$1$. $q=1$: $p=2 (a=k, 4k \leq 40 \Rightarrow k=1..10), p=3 (a=k, 9k \leq 40 \Rightarrow k=1..4), p=4 (a=k, 16k \leq 40 \Rightarrow k=1..2), p=5 (a=k, 25k \leq 40 \Rightarrow k=1), p=6 (a=k, 36k \leq 40 \Rightarrow k=1)$. Total = $10+4+2+1+1 = 18$.$2$. $q=2$: $p=3 (a=4k, 9k \leq 40 \Rightarrow k=1..4), p=5 (a=4k, 25k \leq 40 \Rightarrow k=1)$. Total = $4+1 = 5$.$3$. $q=3$: $p=4 (a=9k, 16k \leq 40 \Rightarrow k=1..2), p=5 (a=9k, 25k \leq 40 \Rightarrow k=1)$. Total = $2+1 = 3$.$4$. $q=4$: $p=5 (a=16k, 25k \leq 40 \Rightarrow k=1)$. Total = $1$.$5$. $q=5$: $p=6 (a=25k, 36k \leq 40 \Rightarrow k=1)$. Total = $1$.Total favorable outcomes = $18 + 5 + 3 + 1 + 1 = 28$.Probability $P = \frac{28}{9880} = \frac{7}{2470}$.Thus,$m = 7, n = 2470$. $\operatorname{gcd}(7, 2470) = 1$.$m + n = 7 + 2470 = 2477$.

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

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