If the ratio of H.M. and G.M. between two num | vedclass

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

Question

(D) We have $H.M. = \frac{2ab}{a + b}$ and $G.M. = \sqrt{ab}$.Given $\frac{H.M.}{G.M.} = \frac{4}{5}$.$\Rightarrow \frac{2ab/(a + b)}{\sqrt{ab}} = \fr

Vedclass · Accepted Answer

$1:4$ and $4:1$

Vedclass · Answer

(D) We have $H.M. = \frac{2ab}{a + b}$ and $G.M. = \sqrt{ab}$.Given $\frac{H.M.}{G.M.} = \frac{4}{5}$.$\Rightarrow \frac{2ab/(a + b)}{\sqrt{ab}} = \frac{4}{5}$ $\Rightarrow \frac{2\sqrt{ab}}{a + b} = \frac{4}{5}$.$\Rightarrow \frac{a + b}{2\sqrt{ab}} = \frac{5}{4}$.Applying componendo and dividendo:$\frac{a + b + 2\sqrt{ab}}{a + b - 2\sqrt{ab}} = \frac{5 + 4}{5 - 4}$ $\Rightarrow \frac{(\sqrt{a} + \sqrt{b})^2}{(\sqrt{a} - \sqrt{b})^2} = \frac{9}{1}$.Taking square root on both sides:$\frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}} = \frac{3}{1}$.Applying componendo and dividendo again:$\frac{(\sqrt{a} + \sqrt{b}) + (\sqrt{a} - \sqrt{b})}{(\sqrt{a} + \sqrt{b}) - (\sqrt{a} - \sqrt{b})} = \frac{3 + 1}{3 - 1}$ $\Rightarrow \frac{2\sqrt{a}}{2\sqrt{b}} = \frac{4}{2} = 2$.Squaring both sides:$\frac{a}{b} = 2^2 = 4$.Thus,$a:b = 4:1$ or $b:a = 1:4$.

If the ratio of $H.M.$ and $G.M.$ between two numbers $a$ and $b$ is $4:5$,then the ratio of the two numbers will be

Similar Questions

If three distinct numbers $a, b, c$ are in $G.P.$ and the equations $ax^2 + 2bx + c = 0$ and $dx^2 + 2ex + f = 0$ have a common root,then which one of the following statements is correct?

For two positive numbers $a$ and $b$,if $a, b$ and $\frac{1}{18}$ are in a geometric progression,while $\frac{1}{a}, 10$ and $\frac{1}{b}$ are in an arithmetic progression,then $16a + 12b$ is equal to $.........$.

If the arithmetic mean of two numbers is $A$ and the geometric mean is $G$,then the numbers are

The product of $n$ positive numbers is unity. Their sum is

Three non-zero real numbers form an $A.P.$ and the squares of these numbers taken in the same order form a $G.P.$ Then the number of all possible common ratios of the $G.P.$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module