Three numbers are selected from the set {3^1, | vedclass

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(B) Let the three numbers in $G.P.$ be $3^a, 3^b, 3^c$ where $1 \le a < b < c \le 20$.For these to be in $G.P.$, the condition is $(3^b)^2 = 3^a \cdot

Vedclass · Accepted Answer

$90$

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(B) Let the three numbers in $G.P.$ be $3^a, 3^b, 3^c$ where $1 \le a For these to be in $G.P.$, the condition is $(3^b)^2 = 3^a \cdot 3^c$, which implies $2b = a + c$.This means $a + c$ must be an even number, which occurs if $a$ and $c$ are both even or both odd.Case $1$: $a$ and $c$ are both odd. There are $10$ odd numbers in the set ${1, 2, \dots, 20}$. We choose $2$ distinct odd numbers for $a$ and $c$ in $^{10}C_2$ ways. Once $a$ and $c$ are chosen, $b$ is uniquely determined as $(a+c)/2$.Number of ways $= ^{10}C_2 = 45$.Case $2$: $a$ and $c$ are both even. There are $10$ even numbers in the set. We choose $2$ distinct even numbers for $a$ and $c$ in $^{10}C_2$ ways. Once $a$ and $c$ are chosen, $b$ is uniquely determined as $(a+c)/2$.Number of ways $= ^{10}C_2 = 45$.Total ways $= 45 + 45 = 90$.

Three numbers are selected from the set ${3^1, 3^2, 3^3, \dots, 3^{20}}$. Find the number of ways these selected numbers can form an increasing Geometric Progression $(G.P.)$.

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