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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $G.P.$

Similar Questions

The first term of an infinite geometric progression is $x$ and its sum is $5$. Then:

If $\alpha, \beta, \gamma$ are the roots of the equation $3x^3-26x^2+52x-24=0$ such that $\alpha, \beta, \gamma$ are in geometric progression and $\alpha < \beta < \gamma$,then $3\alpha + 2\beta + \gamma =$

The sum of the first five terms of the series $3 + 4\frac{1}{2} + 6\frac{3}{4} + \dots$ will be

Let $a_1, a_2, a_3, \ldots$ be a $G.P.$ of increasing positive numbers. Let the sum of its $6^{\text{th}}$ and $8^{\text{th}}$ terms be $2$ and the product of its $3^{\text{rd}}$ and $5^{\text{th}}$ terms be $\frac{1}{9}$. Then $6(a_2 + a_4)(a_4 + a_6)$ is equal to

If $\frac{a + bx}{a - bx} = \frac{b + cx}{b - cx} = \frac{c + dx}{c - dx}$ $(x \ne 0)$,then $a, b, c, d$ are in

Vedclass Test Series

Exam Paper Generator

Online Exam Module