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

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

Vedclass · Accepted Answer

$90$

Vedclass · Answer

(B) Let the three numbers be $3^a, 3^b, 3^c$ where $1 \le a For these to form a $G.P.$,the condition is $(3^b)^2 = 3^a 	imes 3^c$,which implies $2b = a + c$.This means $a$ and $c$ must have the same parity (both even or both odd) so that their sum is even.Case $1$: $a$ and $c$ are both odd.There are $10$ odd numbers in the set $\{1, 2, \dots, 20\}$,which are $\{1, 3, 5, 7, 9, 11, 13, 15, 17, 19\}$.We need to choose $2$ distinct numbers from these $10$ to be $a$ and $c$,which can be done in $^{10}C_2 = 45$ ways.Once $a$ and $c$ are chosen,$b = (a+c)/2$ is uniquely determined and satisfies $a Case $2$: $a$ and $c$ are both even.There are $10$ even numbers in the set $\{1, 2, \dots, 20\}$,which are $\{2, 4, 6, 8, 10, 12, 14, 16, 18, 20\}$.We need to choose $2$ distinct numbers from these $10$ to be $a$ and $c$,which can be done in $^{10}C_2 = 45$ ways.Once $a$ and $c$ are chosen,$b = (a+c)/2$ is uniquely determined and satisfies $a 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 the selected numbers can form an increasing $G.P.$

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