Let G be the geometric mean of two positive n | vedclass

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

Question

(A) $G = \sqrt{ab}$$M = \frac{\frac{1}{a} + \frac{1}{b}}{2} = \frac{a + b}{2ab}$Given that $\frac{1}{M} : G = 4 : 5$,we have $\frac{2ab}{(a + b)\sqrt{

Vedclass · Accepted Answer

$1:4$

Vedclass · Answer

(A) $G = \sqrt{ab}$$M = \frac{\frac{1}{a} + \frac{1}{b}}{2} = \frac{a + b}{2ab}$Given that $\frac{1}{M} : G = 4 : 5$,we have $\frac{2ab}{(a + b)\sqrt{ab}} = \frac{4}{5}$$\Rightarrow \frac{a + b}{2\sqrt{ab}} = \frac{5}{4}$Using Componendo and Dividendo:$\frac{a + b + 2\sqrt{ab}}{a + b - 2\sqrt{ab}} = \frac{5 + 4}{5 - 4}$$\Rightarrow \frac{(\sqrt{a} + \sqrt{b})^2}{(\sqrt{a} - \sqrt{b})^2} = 9$$\Rightarrow \frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}} = 3$ (assuming $a > b$ for positive ratio)$\Rightarrow \sqrt{a} + \sqrt{b} = 3\sqrt{a} - 3\sqrt{b}$$\Rightarrow 4\sqrt{b} = 2\sqrt{a}$ $\Rightarrow \sqrt{\frac{a}{b}} = 2$ $\Rightarrow \frac{a}{b} = 4$Since the ratio $a:b$ can be $1:4$ or $4:1$,and $1:4$ is given in the options,the correct answer is $1:4$.

Let $G$ be the geometric mean of two positive numbers $a$ and $b,$ and $M$ be the arithmetic mean of $\frac{1}{a}$ and $\frac{1}{b}.$ If $\frac{1}{M} : G$ is $4:5,$ then $a:b$ can be

Similar Questions

An $A.P.$,a $G.P.$,and a $H.P.$ have the same first and last terms and the same odd number of terms. The middle terms of the three series are in

If $a, b, c$ are in $A.P.$,then $2^{ax + 1}, 2^{bx + 1}, 2^{cx + 1}$ for $x \ne 0$ are in

If $\frac{a + b}{1 - ab}, b, \frac{b + c}{1 - bc}$ are in $A.P.$,then $a, \frac{1}{b}, c$ are in

In a set of four numbers,the first three are in $G.P.$ and the last three are in $A.P.$ with a common difference of $6$. If the first and last numbers are equal,then the first number is:

The geometric mean of two numbers is $6$ and their arithmetic mean is $6.5$. The numbers are

Vedclass Test Series

Exam Paper Generator

Online Exam Module