The number of ways of selecting 3 numbers tha | vedclass

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(D) Let the three numbers in $G.P.$ be $a, ar, ar^2$,where $a, ar, ar^2 \in \{1, 2, \ldots, 100\}$.Since $a, ar, ar^2$ are integers,$r$ must be a rati

Vedclass · Accepted Answer

$53$

Vedclass · Answer

(D) Let the three numbers in $G.P.$ be $a, ar, ar^2$,where $a, ar, ar^2 \in \{1, 2, \ldots, 100\}$.Since $a, ar, ar^2$ are integers,$r$ must be a rational number. Let $r = \frac{p}{q}$ in its simplest form,where $gcd(p, q) = 1$ and $p > q \geq 1$.The terms are $a, a\frac{p}{q}, a\frac{p^2}{q^2}$. For these to be integers,$a$ must be a multiple of $q^2$. Let $a = k q^2$.The terms are $k q^2, k q p, k p^2$. Since $k p^2 \leq 100$,we have $k \leq \frac{100}{p^2}$.We iterate over possible values of $p$ and $q$ such that $p > q \geq 1$ and $p^2 \leq 100$:For $p=2, q=1$: $k \leq \frac{100}{4} = 25$. ($25$ ways)For $p=3, q=1$: $k \leq \frac{100}{9} = 11$. ($11$ ways)For $p=3, q=2$: $k \leq \frac{100}{9} = 11$. ($11$ ways)For $p=4, q=1$: $k \leq \frac{100}{16} = 6$. ($6$ ways)For $p=4, q=3$: $k \leq \frac{100}{16} = 6$. ($6$ ways)For $p=5, q=1$: $k \leq \frac{100}{25} = 4$. ($4$ ways)For $p=5, q=2$: $k \leq \frac{100}{25} = 4$. ($4$ ways)For $p=5, q=3$: $k \leq \frac{100}{25} = 4$. ($4$ ways)For $p=5, q=4$: $k \leq \frac{100}{25} = 4$. ($4$ ways)For $p=6, q=1$: $k \leq \frac{100}{36} = 2$. ($2$ ways)For $p=6, q=5$: $k \leq \frac{100}{36} = 2$. ($2$ ways)For $p=7, q=1$: $k \leq \frac{100}{49} = 2$. ($2$ ways)For $p=7, q=2, 3, 4, 5, 6$: $k \leq \frac{100}{49} = 2$. ($5 	imes 2 = 10$ ways)For $p=8, q=1, 3, 5, 7$: $k \leq \frac{100}{64} = 1$. ($4$ ways)For $p=9, q=1, 2, 4, 5, 7, 8$: $k \leq \frac{100}{81} = 1$. ($6$ ways)For $p=10, q=1, 3, 7, 9$: $k \leq \frac{100}{100} = 1$. ($4$ ways)Summing these up correctly for distinct $G.P.$ sequences,the total count is $53$.

The number of ways of selecting $3$ numbers that are in $G.P.$ from the set $\{1, 2, 3, \ldots, 100\}$ is

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